No heat is added or removed in an adiabatic process

The change in internal energy in a constant pressure process from temperature $$T_1$$ to $$T_2$$ is equal to $$NC_V(T_2-T_1)$$, where $$C_V$$ is the molar specific heat at constant volume and n is the number of moles of the gas

The change in internal energy of the gas and the work done by the gas are equal in magnitude in an adiabatic process

The internal energy does not change in an isothermal process

Match List I with List II and select the correct answer using the codes given the lists. List I List II P. Boltzmann Constant $$1$$. $$[ML^2T^{-1}]$$ Q. Coefficient of viscosity $$2$$. $$[ML^{-1}T^{-1}]$$ R. Plank Constant $$3$$. $$[MLT^{-3}K^{-1}]$$ S. Thermal conducivity $$4$$. $$[ML^2T^{-2}K^{-1}]$$

P-$$4$$, Q-$$2$$, R-$$1$$, S-$$3$$

P-$$3$$, Q-$$1$$, R-$$2$$, S-$$4$$

P-$$3$$, Q-$$2$$, R-$$1$$, S-$$4$$

P-$$4$$, Q-$$1$$, R-$$2$$, S-$$3$$

MCQ Questions for Class 11 Engineering Physics Kinetic Theory Quiz 9

The temperature (T) of one mole of an ideal gas varies with its volume (V) as

$T = -\alpha { V }^{ 3 }+\ \beta { V }^{ 2 }$ , where

$\alpha$ and

$\beta$ are positive constants. The maximum pressure of gas during this process is

0%
$\dfrac { \alpha \beta }{ 2R }$
0%
$\dfrac { { \beta }^{ 2 }R }{ 4\alpha }$
0%
$\dfrac { (\alpha +\beta )R }{ { 2\beta }^{ 2 } }$
0%
$\dfrac { { \alpha }^{ 2 }R }{ { 2\beta } }$

Explanation

$T=-\alpha V^3+\beta V^2\\ \cfrac{PV}{R}=-\alpha V^2+\beta V^2\\ P=RV(-\alpha V+\beta)$

For maximum pressure

$\cfrac{dP}{dV}=0\\ \Rightarrow -2\alpha V+\beta =0\\ \Rightarrow V=\cfrac{\beta }{2\alpha}\\ P=R[-\alpha(\cfrac{\beta }{2\alpha})^2+\cfrac{\beta }{2\alpha}\times \beta]\\P=R[\cfrac{-\beta^2}{4\alpha}+\cfrac{\beta^2}{2\alpha}]\\P=\cfrac{R\beta^2}{4\alpha}$

When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increase the internal energy of gas is?

0%
$\frac{2}{5}$
0%
$\frac{3}{5}$
0%
$\frac{3}{7}$
0%
$\frac{5}{7}$

Explanation

$f$ =

$∆U$ /

$∆Q$

$nC_{v}∆T$ /

$nC_{p}∆T$

$C_{v}$ /

$C_{p}$

$1$ /

$Y$

$Y$ =

$7$ /

$5$ for ideal gas

$f$ =

$5$ /

$7$

Two identical glass bulbs are interconnected by a thin glass tube at

$0^{\circ}C$ . A gas is filled at N.T.P. in these bulb is placed in ice and another bulb is placed in hot bath, then the pressure of the gas becomes

$1.5\ times$ . The temperature of hot bath will be

0%
$100^{\circ}$
0%
$182^{\circ}C$
0%
$256^{\circ}C$
0%
$546^{\circ}C$

Explanation

Quantity of gas in these bulbs is constant i.e,

Initial no. of moles in both bulb

$=$ final no. of moles

$\Rightarrow n_1+n_2={ n }_{ 1 }^{ ' }+{ n }_{ 2 }^{ ' }$

$\Rightarrow \dfrac{PV}{R(273)}+\dfrac{PV}{R(273)}=\dfrac{1.5PV}{R(273)}+\dfrac{1.5PV}{R(T)}$

$\Rightarrow \dfrac{2}{273}=\dfrac{1.5}{273}+\dfrac{1.5}{T}$

$\Rightarrow T=819K$

$=546°C$

Hence, the answer is

$546°C.$

A vessel of volume $0.3 \ { { m }^{ 3 } }$ contains Helium at $20.0$ . The average kinetic energy per molecule for the gas is:

0%
$6.07\times { 10 }^{ -21 }J$
0%
$7.3\times { 10 }^{ 3 }J$
0%
$14.6\times { 10 }^{ 3 }J$
0%
$12.14\times { 10 }^{ -21 }J$

Explanation

Given temperature of gas

$=20°C$

$=293K$

$\Rightarrow$ Average translational kinetic energy

$=\dfrac{3}{2}KT$

$=\dfrac{3}{2}\times1.38\times10^{-23}\times293$

$=6.07\times10^{-21}J$

Hence, the answer is

$6.07\times10^{-21}J.$

A monoatomic ideal gas undergoes a process in which the ratio of

$P$ to

$V$ at any instant is constant and equal to unity. The molar heat capacity of gas is:

0%
$1.5\ R$
0%
$2.0\ R$
0%
$2.5\ R$
0%
$0$

Explanation

A monoatomic ideal gas undergoes a process in which the ratio of P to V at any instant is constant and equals 1.

$\dfrac{P}{V}=$ constant

$PV^{−1}=$ constant

$PV^{\gamma}=$ constant

$\gamma =−1$

Monatomic gas:

$C_V=\dfrac{3}{2}R$

Molar heat capacity

$=C_V+\dfrac{R}{1−\gamma}$

$=\dfrac{3}{2}R+\dfrac{R}{1−(−1)}$

$=\dfrac{3}{2}R+\dfrac{R}{2}=\dfrac{4R}{2}=2R$

A monoatomic ideal gas, initially at temperature

$T_1$ . is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to temperature

$T_2$ by releasing the piston suddenly. If

$L_1$ and

$L_2$ are the lengths of the gas column before and after expansion respectively, then

$\frac{T_1}{T_2}$ is given by

0%
$(\dfrac{L_1}{L_2})^{2/3}$
0%
$\dfrac{L_1}{L_2}$
0%
$\dfrac{L_2}{L_1}$
0%
$(\dfrac{L_2}{L_1})^{2/3}$

If for a gas

$\dfrac{R}{C_V}=0.67$ , this gas is made up of molecules which are.

0%
Monatomic
0%
Diatomic
0%
Polyatomic
0%
Mixture of diatomic and polyatomic molecules

Explanation

For a gas, we know

$\dfrac{R}{C_V}=\gamma -1$
or

$0.67=\gamma -1$ or,

$\gamma =1.67$
Hence the gas is monatomic.

One Kg of a diatomic gas is at a pressure of

$8\times 10^4$ N/

$m^{2}$ . The density of the gas is

$4$ kg/

$m^{3}$ . The energy of the gas due to its thermal motion will be

0%
$3\times 10^4$ J
0%
$5\times 10^4$ J
0%
$6\times 10^4$ J
0%
$7\times 10^4$ J

Explanation

For a diatomic molecular gas, internal energy per molecule is

$U=\dfrac{5}{2}NK_BT$

$PV=NK_BT$
From equation is (i) and (ii)

$U=\dfrac{5}{2}$ PV
Here,

$P=8\times 10^4N$

$m^{-2}, \rho =4kg$

$m^{-3}$ ,

$m=1$ kg

$\therefore U=\dfrac{5}{2}\times 8\times 10^4\times \dfrac{1}{4}=5\times 10^4$ J

The internal energy of one gram of helium at

$100$ K and one atmospheric pressure is?

0%
$100$ J
0%
$1200$ J
0%
$300$ J
0%
$500$ J

Explanation

The helium molecule is monoatomic and hence its internal energy per molecule is

$\dfrac{3}{2}K_B$ T(where

$k_B$ is the Boltzmann, constant). The internal energy per mole is therefore is

$\dfrac{3}{2}RT$ . One gram of helium is one fourth mole and hence its internal energy is

$\dfrac{1}{4}\times \dfrac{3}{2}R\times 100=300$ J,( By taking the value of R to be approximately

$8$ J

$mol^{-1}$

$K^{-1}$ )

Which one of the following is not an assumption of kinetic theory of gases?

0%
The volume occupied by the molecules of the gas is negligible
0%
The force of attraction between the molecules is negligible
0%
The collision between the molecules are elastic
0%
All molecules have same speed

A vessel of volume V contains a mixture of

$1$ mole of hydrogen and

$1$ mole of oxygen(both considered as ideal). Let

$f_1(v)dv$ denote the fraction of molecules with speed between v and

$(v+dv)$ with

$f_2(v)dv$ , similarly for oxygen. then

0%
$f_1(v)+f_2(v)=f(v)$ obeys the Maxwell's distribution law
0%
$f_1(v), f_2(v)$ will obey the Maxwell's distribution law separately
0%
Neither $f_1(v)$ nor $f_2(v)$ will obey the Maxwell's distribution law
0%
$f_2(v)$ and $f_1(v)$ will be the same

Explanation

The Maxwell-Boltzmann speed distribution function

$\left(N_v=\dfrac{dN}{dv}\right)$ depends on the mass of the gas molecule. [Here, dN is the number of molecules with speeds between v and

$(v+dv)$ ]. The masses of hydrogen and oxygen molecules are different.

A gas mixture consists of

$2$ moles of oxygen and

$4$ moles of Argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is?

0%
$4$ RT
0%
$9$ RT
0%
$11$ RT
0%
$15$ RT

Explanation

For two moles of diatomic oxygen with no vibrational mode.

$U_1=2\times \dfrac{5}{2}RT=5RT$
For four moles of monoatomic Argon,

$U_2=4\times \dfrac{3}{2}RT=6RT$

$\therefore u=u_1+u_1=5RT+6RT=11RT$

If the pressure and the volume of certain quantity of ideal gas are halved, then its temperature

0%
Is doubled
0%
Becomes one-fourth
0%
Remains constant.
0%
Become four times

Explanation

According to ideal gas equation

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$ or

$T_2=\dfrac{P_2V_2}{P_1V_1}$
Here,

$P_1=P, V_1=V, T_1=T, P_2=\dfrac{P}{2}, V_2=\dfrac{V}{2}, T_2=?$

$\therefore T_2=\dfrac{T\left(\dfrac{P}{2}\right)\left(\dfrac{V}{2}\right)}{PV}$ ,

$T_2=\dfrac{T}{4}$ .

The average kinetic energy of

$O_2$ at a particular temperatures is

$0.768$ eV. The average kinetic energy of

$N_2$ molecules in eV at the same temperature is?

0%
$0.0015$
0%
$0.0030$
0%
$0.048$
0%
$0.768$

Explanation

Ideal gas law, says that

$PV = N K T$ .

This means that:

Average Kinetic Energy, $K_{avg} = \dfrac{3}{2} k T$

This is a very important fact, as it tells the fundamentals about the temperature. The absolute temperature of the ideal gas is directly proportional to the average kinetic energy. So, changes in pressure and/or volume will result in changes in temperature too. Also, we can see that the average kinetic energy of the ideal gas

molecule will depend on temperature only and not on the gas nature/mass. Therefore, if the temperature is the same, then the kinetic energy of $O_{2}$ and $N_{2}$ both will be equal.

Hence average K.E. for $N_{2}$ will also be 0.768 eV.

Option D is correct.

Molecular motion shows itself as.

0%
Temperature
0%
Internal energy
0%
Friction
0%
Viscosity

The kinetic energy of

$1$ g molecule of a gas, at normal temperature and pressure, is?

0%
$0.56\times 10^4$ J
0%
$2.7\times 10^2$ J
0%
$1.3\times 10^2$ J
0%
$3.4\times 10^3$ J

Explanation

The kinetic energy of

$1$ g molecule of a gas at temperature

$T=\dfrac{3}{2}$ RT

$=\dfrac{3}{2}\times 8.31\times 273=3.4\times 10^3$ J

Pressure depends on distance as

$P = \dfrac{\alpha}{\beta} exp (- \dfrac{\alpha z}{k \theta})$ , where

$\alpha, \beta$ are constants, z is distance, k is Boltzmann's constant, and

$\theta$ is temperature. The dimensions of

$\beta$ are

0%
$M^0L^0T^0$
0%
$M^{-1}L^{-1}T^{-1}$
0%
$M^0L^2T^0$
0%
$M^{-1}L^{2}T^{2}$

Explanation

$\cfrac{\alpha z}{k\theta}$ is unitless.

$Z\to L$

$\alpha\to MLT^{-2}$

$\cfrac{\alpha}{\beta}=ML^{-1}T{-2}$ (unit of pressure)

$\beta \to L^2$

A vessel contains a mixture consisting of m

$_{1}$ - 7 g of nitrogen (M

$_{1}$ = 28) and m

$_{2}$ = 11 g of carbon dioxide (M

$_{2}$ = 44) at temperature T - 300 K and pressure P

$_{0}$ = 1 atm. The density of the mixture is

0%
$1.46g\ per\ litre$
0%
$2.567 g \ per \ litre$
0%
$3.752 g \ per \ litre$
0%
$4.572 g \ per \ litre$

Explanation

Let the volume occupied

$=V$

By Dalton's law of partial pressure

$\cfrac{P_{nit}}{P_0}=\cfrac{n_{nit}}{n_{nit}+n_{carb}}$

No. of moles of Nitrogen

$\eta_{nit}=\cfrac{M_1}{M_{nit}}=\cfrac{7}{28}=0.25 mol$

No. of moles of carbon

$\eta_{carb}=\cfrac{M_2}{M_{carb}}=\cfrac{11}{4}=0.25 mol$

Thus,

$P_{nit}=P_0\times\cfrac{0.25}{0.25\times0.25}=P_0/2=0.5atm$

From ideal gas equation

$V=\cfrac{nRT}{P}=\cfrac{0.25\times8.314\times290}{0.5\times101325}=0.0119mole$

Total mixture

$m=(7+11)\times 10^{-11}kg$

Thus density s

$P=\cfrac{m}{V}=\cfrac{(7+11)\times 10^{-3}}{0.0119}\approx1.46kg/m^3$

Option A is correct.

A gas performs Q work when it expand at constant pressure. During this process heat absorbed by the gas is 4Q. The average number of degrees of freedom for the gas is:

0%
5
0%
6
0%
4
0%
3.5

Explanation

$Given:$ A gas performs

$Q$ work when it expand at constant pressure. During this process heat absorbed by the gas is

$4Q.$

$Solution:$ In a isobaric process,

$U=4 Q-Q=n C_{v} d T=3 Q$

Work =

$n R dT$ So, U/Work

$=\mathrm{C}_{\mathrm{v}} / \mathrm{R}$

$3=\mathrm{C}_{\mathrm{v}} / \mathrm{R}$

$R=C_{p}-C_{v}$

$3=\dfrac{C_{v}}{C_{p}-C_{v}}$

$\dfrac{C_{p}}{C v}=4 / 3=\gamma$

$\gamma=1+\frac{2}{f}$

where f is degree of freedom

$4 / 3=1+\frac{2}{f}$

$f=6$

$So, the$

$correct$

$option: B$

An ideal gas is initially at

$P_1$ ,

$V_1$ is expanded to

$P_2, V_2$ and then compressed adiabatically to the same volume

$V_1$ and pressure

$P_3$ . If W is the net work done by the gas in the complete process which of the following is true.

0%
$W > 0; P_3 > P_1$
0%
$W < 0; P_3 > P_1$
0%
$W > 0; P_3 < P_1$
0%
$W < 0; P_3 < P_1$

Explanation

Since work is done by the gas as area under the graph is

$-ve$ ( anticlockwise) ie energy is lost by the gas and

$w<0$ when

$P_3>P_1$ .

A vessel contains a mixture consisting of

${m}_{1}=7kg$ of nitrogen

$\left( { M }_{ 1 }=28 \right)$ and

${m}_{2}=11g$ of carbon dioixide

$\left( { M }_{ 2 }=44 \right)$ at temeprature

$T=300K$ and pressure

${ P }_{ 0 }=1\quad atm$ . The density of the mixture is:

0%
$1.446g$ per litres
0%
$2.567g$ per litre
0%
$3.752g$ per litre
0%
$4.572g$ per litre

Explanation

Let V is the volume of the vessel.

Now, let $p_{1}$ and $p_{2}$ be the partial pressure, then using gas law:

$p_{1}V = \dfrac{m_1}{M_1}RT\\$

$p_{2}V = \dfrac{m_2}{M_2}RT,\ p_{0} = p_{1} + p_{2}\\$

$p_{0} = \left(\dfrac{m_1}{M_1} + \dfrac{m_2}{M_2}\right)\dfrac{RT}{V}\\$

$V = \left(\dfrac{m_1}{M_1} + \dfrac{m_2}{M_2}\right)\dfrac{RT}{p_{0}}\\$

$\because \rho_{mix}=\dfrac{(m_{1} + m_{2})}{V}\\$

$\rho_{mix}=\dfrac {(m_1 + m_2)M_1 M_2} {(m_1M_2 + m_2M_1)} \times \dfrac{p_0}{RT}\\$

Substituting values,

$\rho_{mix}=\dfrac {(7 + 11) \times 28 \times 44\times 10^{-3}} {(7 \times 44 + 11\times 28))} \times \dfrac{10^{5}}{8.3 \times 300}\\$

$= 1.446 \ per \ litre$

Option A is correct.

The average degree of freedom per molecule for a gas isThe gas performs 25 J of work when it expands at constant pressure. The heat absorbed by the gas is

0%
75 J
0%
100 J
0%
150 J
0%
125 J

Explanation

Hint: Use First law of thermodynamics

Step 1: Note all the values given.

It is given that aa gas with degree of freedom per molecule of gas

$f = 6$ performs

$W = 25J$ work when it expands st constant Pressure. We know that change in Intenral Energy of gas is given by,

$∆U = \dfrac{f}{2}R∆T$

Where

$R$ is Gas Constant and

$∆T$ is change in temperature.

Therefore,

$∆U = 3R∆T$

Also for Isobaric process, taht is process at constant temperature, Eork done is given by,

$W = P∆V$

$\Rightarrow W = 25J = P∆V$

$\Rightarrow W = R∆T = 25J$ (Since, P∆V = R∆T)

Step 2: Calculate heat absorbed by the gas

By First Law of Thermodynamics, we have

$Q = ∆U + W$

Where

$Q$ is heat given to gas, ∆U is change in Internal Energy and W is work done by gas. We have

$Q = 3R∆T + R∆T$

$\Rightarrow Q = 4R∆T$

$\Rightarrow Q = 4 \times 25J$

$\Rightarrow Q = 100J$

Therefore, heat absorbed by gas is, $Q = 100J$

Option B is correct.

A gas thermometer measures the temperature from the variation of pressure of a sample of gas. If the pressure measured at the melting point of lead is

$2.20\ times$ the pressure measured at the triple point of water find the melting point of lead.

0%
$600K$
0%
$420K$
0%
$790K$
0%
$510K$

Explanation

$\textbf{Correct option: Option (A).}$

$\textbf{Explanation:}$

Given, the pressure at melting point is 2.20 times the pressure at triple point of water.

${{P}_{MP}}=2.20\times {{P}_{TP}}$

We know that, Melting point, $MP=\dfrac{{{P}_{MP}}}{{{P}_{TP}}}\times 273.16K$

$\Rightarrow MP=\dfrac{2.20\times {{P}_{TP}}}{{{P}_{TP}}}\times 273.16\approx 600K$

$\textbf{So, correct option is (A).}$

For a ideal gas.

0%
The change in internal energy in a constant pressure process from temperature $T_1$ to $T_2$ is equal to $NC_V(T_2-T_1)$ , where $C_V$ is the molar specific heat at constant volume and n is the number of moles of the gas
0%
The change in internal energy of the gas and the work done by the gas are equal in magnitude in an adiabatic process
0%
The internal energy does not change in an isothermal process
0%
No heat is added or removed in an adiabatic process

Explanation

Change in internal energy of a gas

$=\dfrac{f}{2}nR\triangle T$

In an isothermal process,

$\triangle T=0$

$\Rightarrow$ Change in internal energy

$=0$

So, option (C) is correct.

Now,

$C_v=\dfrac{f}{2}R$

$\Rightarrow \triangle u=n\left(C_v\right) \triangle T=nC_v\left(T_2-T_1\right)$

So, option (A) is also correct.

First law of thermodynamics :

$dQ=dv+dw$

In an adiabatic process,

$dQ=0$ [ option (D) is also correct. ]

$\Rightarrow dv=-dw$

$\Rightarrow \left| dv \right| =\left| dw \right|$

So, option (B) is also correct.

Hence, this question has multiple answers.

Match List I with List II and select the correct answer using the codes given the lists.

List I	List II
P. Boltzmann Constant	$1$ . $[ML^2T^{-1}]$
Q. Coefficient of viscosity	$2$ . $[ML^{-1}T^{-1}]$
R. Plank Constant	$3$ . $[MLT^{-3}K^{-1}]$
S. Thermal conducivity	$4$ . $[ML^2T^{-2}K^{-1}]$

0%
P- $3$ , Q- $1$ , R- $2$ , S- $4$
0%
P- $3$ , Q- $2$ , R- $1$ , S- $4$
0%
P- $4$ , Q- $2$ , R- $1$ , S- $3$
0%
P- $4$ , Q- $1$ , R- $2$ , S- $3$

Explanation

$E=\cfrac{3}{2}KT\\ Boltzman constant\rightarrow[K]=\cfrac{[ML^2T^{-2}]}{[\theta]}=[ML^2T^{-2}K^{-1}]$

Coefficient of viscosity of

$\rightarrow \eta=\cfrac{F}{A/dv/dx}=\cfrac{MLT^{-2}}{L^2\times T^{-1}}=[ML^{-1}T^{-1}]$

Planck's constant

$\rightarrow ML^2T^{-1}$

Dimension formula of thermal conductivity

Thermal conductivity is measured in

$(wall/m.K)\\=MLT^{-3}K^{-1}$

An insulated box containing monatomic ideal gas of molar mass M is moving with a uniform speed v. The box suddenly stops and consequently the gas acquires a new temperature. Calculate the change in the temperature of the gas. Neglect heat absorbed by the box.

0%
$\Delta T=2\dfrac{Mv^2}{3R}$ .
0%
$\Delta T=\dfrac{Mv^2}{3R}$ .
0%
$\Delta T=3\dfrac{Mv^2}{3R}$ .
0%
$\Delta T=4\dfrac{Mv^2}{3R}$ .

Explanation

An insulated Box containing monoatomic ......... neglect heat absorbed by box

molar mass

$=M$

speed

$=v$

Initial energy

$=KE=\dfrac{1}{2}MV^{2}$

Final energy = Temperature energy (or

$PE$ )

$=\dfrac{3}{2}RT$

For ideal conditions,

$\dfrac{1}{2}, mv^{2}=\dfrac{3}{2}RT$

$\Delta T=\dfrac{mv^{2}}{3R}$

A gas molecule of mass

$M$ at the surface of the earth has kinetic energy equivalent to

${0}^{o}C$ . If it were to go up straight without colliding with any other molecules, how high it would rise? Assume that the height attained is much less than radius of the earth. (

${k}_{B}$ is Boltzmann constant)

0%
zero
0%
$\cfrac { 273{ k }_{ B } }{ 2Mg }$
0%
$\cfrac { 546{ k }_{ B } }{ 3Mg }$
0%
$\cfrac { 819{ k }_{ B } }{ 2Mg } \quad$

An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes

$5.66$ V while its temperature falls to

$T/2$ . How many degrees of freedom do the gas molecules have?

0%
7
0%
$5$ .
0%
6
0%
8

Explanation

Adiabatic equation of perfect gas is given as

$TV^{r-1}=$ constant

$m=T_{1}V_{1}^{(r-1)}=T_{2}V_{2}^{(r-1)}$

$T_{1}=T_{1}V_{1}=V_{1}V_{2}=5.66\ V$

and

$T_{2}=\dfrac{T}{2}$

$TV^{r-1}=\dfrac{T}{2}(5.66\ V)^{r-1}$

$2=5.66^{r-1}$

Taking

$\log$ on both sides

$(r-1)\log 5.66=\log 2(r-1)$

$r=\dfrac{\log 2}{\log 5.66}=1+0.3010/0.75$

$r=1+0.4$

$r=1.4$ for

$r=1.4$ Agree of freedom

$=5$

$m$ grams of a gas of molecular weight

$M$ is flowing in an insulated tube with velocity

$v$ . If the system is suddenly stopped then the rise in its temperature will be (γ= ratio of specific heats, R=universal gas constant, J=Mechanical equivalent of heat)

0%
$\dfrac {Mv^{2}(\gamma - 1)}{2RJ}$
0%
$\dfrac {Mv^{2}\gamma}{2RJ}$
0%
$\dfrac {v^{2}}{2Js}$
0%
$\dfrac {Mv^{2}(\gamma - 1)}{2MRJ}$

Two monatomic ideal gases

$1$ and

$2$ of molecular masses

$m_1$ and

$m_2$ respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas

$1$ to that in gas

$2$ is given by.

0%
$\sqrt{\dfrac{m_1}{m_2}}$
0%
$\sqrt{\dfrac{m_2}{m_1}}$
0%
$\dfrac{m_1}{m_2}$
0%
$\dfrac{m_2}{m_1}$

Explanation

$V_{rm}=\sqrt {\dfrac {rRT}{m}}$

i.e

$vr\dfrac {1}{\sqrt m}$

which given

$v-1\times \dfrac {1}{\sqrt {m_1}}$ and

$v_2\times \dfrac {1}{\sqrt {m_1}}$

$\dfrac {v_1}{v_2}=\sqrt {\dfrac {m_2}{m_1}}$

Two cylinders contain the same amount of ideal monatomic gas. The same amount of heat is given to two cylinders. If the temperature rise in cylinder A is

$T_0$ then temperature rise in cylinder B will be :

0%
$\dfrac{4}{3}T_0$
0%
$2T_0$
0%
$\dfrac{T_0}{2}$
0%
$\dfrac{5}{3}T_0$

Explanation

$Q=nC_vT_0$

$Q_A=nC_p\triangle T_A$

$Q_B=nC_v\triangle T_B$

$Q_A=Q_B$

$nC_pT_0=nC_v\triangle T$

$\triangle T=T_0\times \dfrac{C_p}{C_v}$

$=T_0\times \dfrac{5}{3}.$

Hence, the answer is

$\dfrac{5}{3}T_0.$

One kg of a diatomic gas is at a pressure of

$8\times {10}^{4}N/{m}^{2}$ . The density of gas is

$4kg/{m}^{2}$ . What is the energy of the gas due to its thermal motion?

0%
$6\times {10}^{4}J$
0%
$7\times {10}^{4}J$
0%
$3\times {10}^{4}J$
0%
$5\times {10}^{4}J$

Ratio of

$C_P$ and

$C_V$ of gas '

$X$ ' is 1.The number of atoms of the gas '

$X$ ' present in

$11.2$ litres of at NTP will be:

0%
$6.02\times 10^{23}$
0%
$5.61\times 10^{23}$
0%
$3.01\times 10^{23}$
0%
$2.01\times 10^{23}$

Explanation

$\dfrac{C_p}{C_v} = \gamma = 1.4$

Hence, the gas is diatomic.

Number of atoms

$=$ atoms in one molecule of gas

$\times$ mole

$\times N_A$

$= 2 \times \dfrac{V}{22.4} \times N_A$

$= 2 \times \dfrac{11.2}{22.4}\times 6.02\times 10^{23} = 6.02\times 10^{23}$

Valency of carbon is

44. From this, we understand that there are ____chemical bond/bonds between the carbon atom and one oxygen atom in the compound-carbon dioxide.

0%
$1$
0%
$2$
0%
$3$
0%
$4$

If temperature of body increases by 10%, then increase in radiated energy of the body is :

0%
10 %
0%
40 %
0%
46 %
0%
1000 %

The average thermal energy of a oxygen atom at room temperature

$(27^{o}C)$

0%
$4.5\times {10}^{-21}{J}$
0%
$6.2 \times {10}^{-21}{J}$
0%
$3.4 \times {10}^{-21}{J}$
0%
$1.8 \times {10}^{-21}{J}$

Explanation

Boltzman's constant is given as

$k=1.38 \times 10^{-23} J/k$

Average kinetic energy will be

$KE=\dfrac{FkT}{2}$

Where

$F$ is degree of freedom which will be

$3$ because

$O$ is a mono-atomic molecule.

Temperature,

$T=27+273=300K$

So energy=

$\dfrac{3 \times 1.38 \times 10^{-23} \times 300}{2}=6.2\times 10^{-21} Joule$

Option B is correct.

The amount of heat energy required to raise the temperature of

$1\ g$ of helium in a container of volume

$10L$ , from

$T_{1}\ K$ to

$T_{2}\ K$ is (

$N_{a} =$ Avogadros number,

$k_{B}=$ Boltzmann constant)

0%
$\dfrac { 3 }{ 2 } { N }_{ a }{ k }_{ B }\left( { T }_{ 2 }-{ T }_{ 1 } \right)$
0%
$\dfrac { 3 }{ 7} { N }_{ a }{ k }_{ B }\left( { T }_{ 2 }-{ T }_{ 1 } \right)$
0%
$\dfrac { 3 }{ 4 } { N }_{ a }{ k }_{ B }\left( { T }_{ 2 }-{ T }_{ 1 } \right)$
0%
$\dfrac { 3 }{ 8 } { N }_{ a }{ k }_{ B }\left( { T }_{ 2 }-{ T }_{ 1 } \right)$

Explanation

The process is being completed at a constant volume because the volume of the container is fixed.

so the heat energy required will be

$Q=nC_v \Delta T$

Where

$C_v$ is specific heat at constant volume and

$n$ is number of moles given by

$n=\dfrac{1gm}{4gm/mole}=\dfrac{1}{4}mole$

and the specific heat is given as

$C_v=\dfrac{fN_ak_B}{2}$

The degree of freedom,

$f$ has value

$3$ for a mono-atomic gas.

$N_a$ is Avogadro Number and

$k_B$ is Boltzman's constant

putting all above values in the expression for

$Q$ we get

$Q=\dfrac{3}{8}N_ak_B \Delta T=\dfrac{3}{8}N_ak_B(T_2 -T_1)$

Option D is correct.

The kinetic energy of

$1g$ molecule of a gas at normal temperature and pressure is :

0%
$1.3$ x $10^2 J$
0%
$2.7$ x $10^2 J$
0%
$0.56$ x $10^4 J$
0%
$3.4$ x $10^3 J$

Explanation

There is no information about gas so the translational kinetic energy is given as,

$K = \dfrac{3}{2}RT$

$= \dfrac{3}{2} \times 8.31 \times \left( {273 + 0} \right)$

$= 3.4 \times {10^3}\;{\rm{J}}$

Thus, the kinetic energy of a gas molecule is $3.4 \times {10^3}\;{\rm{J}}$ .

In an experiment, 1.35 mol of oxygen (O2) are heated at constant pressure starting at 11.0ºC. How much heat must be added to the gas to double its volume?

0%

$1.12\times 10^4 J$
0%

$1.40\times 10^4$
0%
$$2.12\times 10^4$$J
0%
$$3.12×10^4 J$$

Explanation

Since oxygen

$(O_2)$ is a diatomic gas

$\therefore C_v=5/2R$

So,

$C_p=C_v+R\\ \quad=5/2R+R\\ \quad =7/2R$

At constant pressure

If we double the volume the temperature will be doubled.

$T_i=11.0°C=(11.0+273)K=284K$

So,

$T_p=2(284)K=568K$
To obtain heat,

$Q=nC_p(T_p-T_i)$

$Q=(1.35mol)(7/2R)(508K-284K)\\Q=1.12\times10^4J$

By the ideal gas law, the pressure of

$0.60$ moles

${NH}_{3}$ gas in a

$3.00\ L$ vessel at

${25}^{o}C$ is, given that

$R=0.082\ L$ atm

${mol}^{-1}{k}^{-1}$ :

0%
$48.9\ atm$
0%
$4.89\ atm$
0%
$0.489\ atm$
0%
$489\ atm$

Explanation

Pressure is given as

$P=\dfrac{nRT}{V}=\dfrac{0.6\times .082\times (25+273)}{3}=4.89atm$

Option B is correct.

Mean kinetic energy (or average energy) per gm. of a molecule of a monoatomic gas is given by :

0%
$\dfrac{3RT}{2}$
0%
$\dfrac{kT}{2}$
0%
$\dfrac{RT}{3}$
0%
$\dfrac{3kT}{2}$

RMS velocity of an ideal gas at

$27^o C$ is

$500{m/s}$ , Temperature is increased four times, RMS velocity will become.

0%
$1000{m/s}$
0%
$560{m/s}$
0%
$2000m/s$
0%
None of these

Explanation

$V_{rms}=\sqrt{\cfrac{3RT}{M}}=500$

${m/s}$

$V_{rms}^{\prime}=\sqrt{\cfrac{3RT^{\prime}}{M}} =\sqrt{\cfrac{3R(4T)}{M}}$

$V_{rms}^{\prime}=2\sqrt{\cfrac{3RT}{M}}=2V_{rms}=1000$

${m/s}$

If the total number of

$H_2$ molecules is double of the

$O_2$ molecules then the ratio of total kinetic energies pf

$H_2$ to that of

$O_2$ at 300 K is :-

0%
1 : 1
0%
1 : 2
0%
2 : 1
0%
1 : 3

Explanation

Kinetic energy for a single molecule is given as

$E=\dfrac{3}{2}kT$

where

$k$ is Boltzman's constant so for

$2n$ molecules of Hydrogen the kinetic energy will be

$E_{H_2}=2n(1.5kT)$

And that for

$n$ molecules of

$O_2$ will be

$E_{O_2}=n(1.5kT)$

ratio of energi

$es$ will be

$E_{H_2} :E_{O_2}=2:1$

Option C is correct.

Ideal monoatomic gas is taken through a process

$dQ =2dU$ . What is the molar heat capacity for the process? ( where dQ is heat supplied and dU is change in internal energy)

0%
2R
0%
3R
0%
4R
0%
5R

Explanation

$d Q= nCp dT$

$Cp-$ heat capacity of constant pressure

$dU=nCvdT$

$Cv-$ heat capacity at constant volume

$n-$ number of moles

$dQ= 2d U$

$nCpdT= 2nCvdT$

$Cp= 2Cv$

$= 2 \left( \dfrac{R}{y-1} \right)$

(For mono atomic

$y= 5/3$ )

$=\dfrac{2R}{\dfrac{5}{3}-1}$

$Cp= 3R$

The relation between the ratio of specific heats

$(\gamma)$ of gas and degree of freedom

$'f'$ will be :

0%
$\gamma=f+2$
0%
$\dfrac{1}{\gamma}=\dfrac{1}{f}+\dfrac{1}{2}$
0%
$f=\dfrac{2}{\gamma-1}$
0%
$f=2(\gamma-1)$

Three particles are situated on a light and rigid rod along Y-axis as shown in the figure. If the system is rotating with angular velocity of

$2 rad/sec$ about X axis, then the total kinetic energy of the system is :

0%
$92 J$
0%
$184 J$
0%
$276 J$
0%
$46 J$

The total kinetic energy of 1 mole of

${ N }_{ 2 }$ at

$27^oC$ will be approximately:

0%
$1500J$
0%
$1500$ calorie
0%
$1500$ kilo calorie
0%
$1500$ erg

Explanation

According to law of Equipartition of energy, a molecule can have

$\dfrac{nRT}{2}$ energy per degree of freedom.

Here given molecule

$N_2$ is diatomic molecule and we know, there are five degree of freedom of a diatomic molecule ( three translational and two rotational ).

So, total kinetic energy =

$\dfrac{5nRT}{2}$

Here,

$n$ is number of mole. given,

$n = 1$

$R$ is universal gas constant i.e.,

$R = 2 Cal/mol.K$

and T is temperature in Kelvin. given,

$T = 27°C = (27 + 273)K = 300K$

now, total kinetic energy =

$5/2 × 1 × 2 × 300$

$5 × 300 = 1500 Cal$

The total Kinetic energy of

$1\ mole$ of

${N}^{}_{2}$ at

$27^{o}_{}{C}$ will be approximately :-

0%
$1500\ J$
0%
$15633\ cal$
0%
$1500\ kcal$
0%
$1500\ erg$

Explanation

For

$n$ mole of any gas the total kinetic energy is given as

$E=\dfrac{3}{2}nRT$

Where

$R$ is gas constant having value

$8.31J/mole-K$ or

$8.31\times 4.18 cal /mole-K=34.74\text{Cal per mole per Kelvin}$

$T$ is temperature in Kelvin which is

$T=27+273=300K$

So putting all values we get

$E=1.5\times 1 \times 34.74\times 300=15633Calorie$

$3$ mole of gas ''X" and

$2$ moles of gas "Y" enters from end "P" and "Q" of the cylinder respectively. The cylinder has the area of cross section , shown as
under
The length of the cylinder is

$150cm$ . The gas "X" intermixes with gas "Y" at the point . If the molecular weight of the gases X and Y is

$20$ and

$80$ respectively, then what will be the distance of point A from Q?

0%
$75cm$
0%
$50cm$
0%
$37.5$
0%
$90cm$

Explanation

$\begin{array}{l} \frac { { rx } }{ { ry } } =\frac { { { w_{ x } } } }{ { { n_{ y } } } } \sqrt { \frac { { { M_{ y } } } }{ { { M_{ x } } } } } \\ =\frac { 3 }{ 2 } \sqrt { \frac { { 80 } }{ { 20 } } } =\frac { 3 }{ 1 } =3:1 \\ \therefore \frac { { dis\tan ce\, \, travelled\, \, by\, \, gas\, \, X } }{ { dis\tan ce\, \, travelled\, \, by\, \, gas\, \, Y } } =3:1 \\ \therefore dis\tan ce\, \, of\, \, A\, \, from\, \, Q=\frac { { 150 } }{ 3 } =50\, \, cms \end{array}$

Hence, OPtion

$B$ is correct.

How many degrees of freedom the gas molecules have if under STP the gas density

$\rho = 1.3 kg/m^3$ and the velocity of sound propagation in it is

$330 ms^{-1}$ ?

0%
$3$
0%
$5$
0%
$7$
0%
$8$

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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