MCQ Questions for Class 11 Medical Chemistry Thermodynamics Quiz 1

A chemical reaction is spontaneous at 298 K but non-spontaneous at 350 K. Which one of the following is true for the reaction?

0%
$\Delta G\quad \Delta H\quad \Delta S\\ -\quad \quad -\quad \quad +$
0%
$\Delta G\quad \Delta H\quad \Delta S\\ +\quad \quad +\quad \quad +$
0%
$\Delta G\quad \Delta H\quad \Delta S\\ -\quad \quad +\quad \quad -$
0%
$\Delta G\quad \Delta H\quad \Delta S\\ +\quad \quad -\quad \quad +$
0%
$\Delta G\quad \Delta H\quad \Delta S\\ -\quad \quad -\quad \quad -$

Which of the following statements regarding Gibb's energy change is correct?

0%
If $\Delta$ G is negative (< 0), the process is non-spontaneous.
0%
If $\Delta$ G is positive (> 0), the process is non-spontaneous.
0%
If $\Delta$ G is negative (< 0), the process is spontaneous.
0%
If $\Delta$ G is positive (> 0), the process is equilibrium.

Explanation

The energy available for a system at some conditions and by which useful work can be done is Gibb's energy (G).

$G=H-TS$

For spontaneous process,

$\Delta G<0$

For non-spontaneous process,

$\Delta G>0$

For equilibrium process,

$\Delta G=0$

$\triangle H$ = - 336.2 kcal. What is the value of

$\triangle U$ approximately at 300 K for the same reaction (R = 2 cal

$degre{ e }^{ -1 }$

$mo{ l }^{ -1 }$ )?

0%
- 320.0 kcal
0%
- 335.0 kcal
0%
- 337.2 kcal
0%
- 353.0 kcal

Properties of substances like pressure, temperature and density, in thermodynamic coordinates are

0%
path function
0%
point function
0%
cyclic function
0%
real function

The INCORRECT match in the following is :

0%
$\Delta G^0 < 0,K< 1$
0%
$\Delta G^0 < 0,K= 1$
0%
$\Delta G^0 > 0,K< 1$
0%
$\Delta G^0 < 0,K> 1$

Explanation

We know,

$\Delta G=\Delta G^0+RT\ lnK$

At equilibrium,

$\Delta G=\Delta G^0+RT\ lnK=0$

$\Rightarrow \Delta G^0=−RT\ lnK$

$−RT\ lnK<0$

$RT\ lnK>0$

$lnK>0$

$\Rightarrow K>1$

Which of the following property is not a thermodynamic property of the system?

0%
pressure
0%
temperature
0%
specific volume
0%
heat

Which heat depends on the direction of current?

0%
Joule heat
0%
Peltier heat
0%
Thompson effect
0%
None of these

An ideal gas undergoes a cyclic process as shown in Figure.

$\Delta U_{BC}=-5$ kJ

$mol^{-1}$ ,

$q_{AB}=2$ kJ

$mol^{-1}$

$W_{AB}=-5$ kJ

$mol^{-1}$ ,

$W_{CA}=3$ kJ

$mol^{-1}$

Heat absorbed by the system during process CA is:

0%
$+18$ kJ $mol^{-1}$
0%
$-18$ kJ $mol^{-1}$
0%
$-5$ kJ $mol^{-1}$
0%
$+5$ kJ $mol^{-1}$

Explanation

$\Delta U _{AB} = q_{AB}+W_{AB}=2+(-5)=-3 kJ/mol$

$\Delta U _{BC} = -5 kJ/mol$

For cyclic process,

$\Delta U = 0$

$\Delta U _{AB}+ \Delta U _{BC} + \Delta U_{CA} = 0$

$\Delta U_{CA} = -\Delta U _{AB}- \Delta U _{BC}$

$\Delta U_{CA} = - (-3 )- (-5)=8 kJ/mol$

$\Delta U_{CA} =q_{CA}+W_{CA}$

$8 =q_{CA}+3$

$q_{CA}=+5kJ/mol$

Heat absorbed has positive sign.

The combustions of benzene (l) gives

$CO_2(g)$ and

$H_2O(l)$ . Given that heat of combustion of benzene at constant volume is

$-3263.9 \ kJ\, mol^{-1}$ at

$25^oC$ , heat of combustion (in

$kJ\,mol^{-1}$ ) of benzene at constant pressure will be:

$(R =8.314JK^{-1}\,mol^{-1})$

0%
$+3260$
0%
$-3267.6$
0%
$+4152.6$
0%
$-452.46$

Explanation

$C_6H_6(l) + \dfrac{15}{2} O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)$

$\Delta H = \Delta U + \Delta n RT$

$=-3263.9 + \dfrac {( 6- 7.5) \times 8.314 \times 298}{1000}$

$=-3263.9 - 3.7$

$= -3267.6 \ kJ$

Hence, option

$B$ is correct.

The standard Gibbs energy for the given cell reaction in

$kJ$

${mol}^{-1}$ at

$298K$ is:

$Zn(s)+{Cu}^{2+}(aq)\rightarrow {Zn}^{2+}(aq)+Cu(s)$

${E}^{o}_{cell}=2V$ at

$298K$
(Faraday's constant,

$F=96000\ C{mol}^{-1}$ )

0%
$-384$
0%
$-192$
0%
$192$
0%
$384$

Explanation

We know,

$\Delta G^o = -nFE^o_{cell}$

Since 2 electrons are involved so

$n=2$ .

$\Delta G = -nFE^o_{cell} = -2 \times 96000 \times 2 =-384000\ J= -384\ kJ$

A process will be spontaneous to all temperatures if:

0%
$\Delta H > 0$ and $\Delta S < 0$
0%
$\Delta H < 0$ and $\Delta S > 0$
0%
$\Delta H > 0$ and $\Delta S > 0$
0%
$\Delta H < 0$ and $\Delta S < 0$

Explanation

Solution:- (B)

$\Delta H < 0$ and

$\Delta S > 0$

$\Delta G = \Delta H - T \Delta S$

For reaction to be spontaneous process at all temperature,

$\Delta G < 0$ and it is possible when

$\Delta H < 0$ and

$\Delta S > 0$ because it gives

$\Delta G < 0$ .

Compute the heat of formation of liquid methyl alcohol (in kJ mol

$^{ -1 }$ ) using the following data. The heat of vaporisation of liquid methyl alcohol is 38 kJ mol

$^{ -1 }$ .

The heat of formation of gaseous atoms from the elements in their standard states:

$H =$ 218 kJ mol

$^{ -1 }$ .,

$C = 715$ kJ mol

$^{ -1 }$ . ,

$O = 249$ kJ mol

$^{ -1 }$ .

Average bond energies:

$C-H = 415$ kJ mol

$^{ -1 }$ .

$C-O = 356$ kJ mol

$^{ -1 }$ .

$O-H = 463$ kJ mol

$^{ -1 }$ .

0%
$\Delta H = -266$ kJ mol $^{ -1 }$ .
0%
$\Delta H = +266$ kJ mol $^{ -1 }$ .
0%
$\Delta H = -190$ kJ mol $^{ -1 }$ .
0%
None of these

Explanation

$CH_3OH_{(l)}\longrightarrow CH_3OH_{(g)}; \Delta H = 38$ kJ mol

$^{ -1 }$ . .....(1)

$\dfrac{ 1 }{ 2 }$

${H_2}_{(g)}$

$\longrightarrow H_{(g)}; \Delta H = 218$ kJ mol

$^{ -1 }$ . .....(2)

$C(graphite)\longrightarrow C_{(g)}; \Delta H = 715$ kJ mol

$^{ -1 }$ . .....(3)

$\frac{ 1 }{ 2 }{O_2}_{(g)}\longrightarrow O_{(g)}; \Delta H = 249$ kJ mol

$^{ -1 }$ . .....(4)

$C-H_{(g)}\longrightarrow C_{(g)} + H_{(g)}; \Delta H = 415$ kJ mol

$^{ -1 }$ . .....(5)

$C-O_{(g)}\longrightarrow C_{(g)} + O_{(g)}; \Delta H= 356$ kJ mol

$^{ -1 }$ . .....(6)

$O-H_{(g)}\longrightarrow O_{(g)} + H_{(g)}; \Delta H = 463$ kJ mol

$^{ -1 }$ . .....(7)

$C(s) + 2{H_2}_{(g)} + \frac{ 1 }{ 2 }{O_2}_{(g)}\longrightarrow CH_3OH_{(g)}$ .....(8)

The following energies are required:

$2{H_2}_{(g)}\longrightarrow 4H_{(g)}$

$C(graphite)\longrightarrow C_{(g)}$

$\frac{1 }{ 2 }{O_2}_{(g)}\longrightarrow O_{(g)}$

The total energy required is

$(218\times 4) + 175 + 249 = 1836\ kJ$ .....From equation (5), (6) and (7)

The following bonds are formed:

Three C-H bonds, one C-O bond, and O-H bond

The total energy released is

$(3\times 415) + 356 + 463 = 2064\ kJ$

Net energy released for

$CH_3OH_{(l)} = 228 + 38 = 266\ kJ$ .... From eq (1) and (2)

Therefore,

$\Delta H_fCH_3OH_{(l)} = -266$ kJ mol

$^{ -1 }$

Hence, the correct option is

$A$

For the process :

$\\ \mathrm{H}_{2}\mathrm{O}(l)$ (

$1$ bar,

$373$

$\mathrm{K}$ )

$\rightleftharpoons \mathrm{H}_{2}\mathrm{O}(\mathrm{g})$ (

$1$ bar,

$373 \mathrm{K}$ ),

the correct set of thermodynamic parameters is

0%
$\Delta \mathrm{G}=0,\Delta \mathrm{S}=+ ve$
0%
$\Delta \mathrm{G}=0, \Delta S=-ve$
0%
$\Delta \mathrm{G}=+ve, \Delta \mathrm{S}=0$
0%
$\Delta \mathrm{G}=- ve, \Delta \mathrm{S}=+ ve$

Explanation

$\mathrm{H}_{2}\mathrm{O}(l)$ ( 1 bar, 373

$\mathrm{K}$ )

$\rightarrow \mathrm{H}_{2}\mathrm{O}(\mathrm{g})$ ( 1 bar, 373 )

As this process is in equilibrium so

$\Delta \mathrm{G}=0$

Also, here water changes from liquid to more random state gas so entropy of system increases.

$\Delta \mathrm{S}=+$ ve

A gas mixture consists of

$2$ moles of

$O_2$ and

$4$ moles of Ar at temperature

$T$ . Neglecting all vibrational modes, the total internal energy of the system is

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

Explanation

Total internal energy

$\nu = \dfrac{f_{O_2}}{2} n_{O_2}RT+\dfrac{f_{Ar}}{2}n_{A_r}$

$\nu = \dfrac{5}{2}\times 2RT +\dfrac{3}{2}\times 4RT$

$\nu =5RT + 6RT$

$\nu = 11RT$

For a given reaction,

$\triangle H = 35.5\ kJmol^{-1}$ and

$\triangle S = 83.6\ JK^{-1} mol^{-1}$ . The reaction is spontaneous at: (Assume that

$\triangle H$ and

$\triangle S$ do not vary with temperature)

0%
$T > 298\ K$
0%
$T < 425\ K$
0%
$T > 425\ K$
0%
All temperatures

Explanation

The reaction will be spontaneous when

$\displaystyle \Delta G$ = negative

$\displaystyle \Delta G = \Delta H - T \Delta S$

$\Delta H$ <

$T\Delta S$

$\displaystyle T >\dfrac { \Delta H}{ \Delta S}$

$\displaystyle T = \dfrac {35.5 \: kJ/mol\times10^3\ J/kJ}{83.6 \: J/K/mol}$

$\displaystyle T =425$

The reaction is spontaneous when

$\displaystyle T > 425\ K$

The following two reactions are known

$Fe_2O_{3(s)} + 3CO_{(g)}\rightarrow 2Fe(s) + 3CO_{2(g)}$ ;

$\Delta \ H = -26.8\ kJ$

$FeO_{(s)} + CO_{(g)}\rightarrow Fe_{(s)} + CO_{2(g)}\Delta H = -16.5\ kJ$

The value of

$\Delta\ H$ for the following reaction is:

$Fe_2O_{3(s)} + CO_{(g)}\rightarrow 2FeO_{(s)} + CO_{2(g)}$

0%
+ 10.3 kJ
0%
- 43.3 kJ
0%
- 10.3 kJ
0%
+ 6.2 kJ

Explanation

Given :
(I) Fe

$_2$ O

$_{3(s)}$ + 3CO

$_{(g)}$

$\rightarrow$ 2Fe(s) + 3CO

$_{2(g)}$ ;

$\Delta$

$H = -26.8 kJ$

(II) FeO

$_{(s)}$ + CO

$_{(g)}$

$\rightarrow$ Fe

$_{(s)}$ + CO

$_{2(g)}$ ;

$\Delta$

$= -16.5 kJ$

On multiplying equation (II) with 2, we get

(III) 2FeO

$_{(s)}$ + 2CO

$_{(g)}$

$\rightarrow$ 2Fe

$_{(s)}$ + 2CO

$_{2(g)}$ ;

$\Delta$

$H = -33 kJ$

On substracting equation (IIII) from, I we get

(IV) Fe

$_2$ O

$_{3(s)}$ + CO

$_{(g)}$

$\rightarrow$ 2FeO

$_{(s)}$ + CO

$_{2(g)}$ ;

$\Delta$

$H = -26.8 - (-33) kJ = + 6.2 kJ$

The value of

$\Delta$ H for the reaction (IV) is +

$6.2 kJ.$

Which of the following is the property of a system

0%
pressure and temperature
0%
internal energy
0%
volume and density
0%
all of the above

Consider the following processes:

$\Delta H\left({kJ}/{mol}\right)$

$\dfrac{1}{2} A \rightarrow B$

$+150$

$3B \rightarrow 2C + D$

$-125$

$E + A \rightarrow 2D$

$+350$

For

$B + D \rightarrow E + 2C, \Delta H$ will be:

0%
$525\ {kJ}/{mol}$
0%
$-175\ {kJ}/{mol}$
0%
$-325\ {kJ}/{mol}$
0%
$325\ {kJ}/{mol}$

Explanation

The processes are as shown below.

$\Delta H$

$\displaystyle\frac{1}{2}A \longrightarrow B$

$+150$ ............(1)

$3B \longrightarrow 2C + D$

$-125$ ...............(2)

$E + A \longrightarrow 2D$

$+350$ ..................(3)
________________________

$B + D \longrightarrow E + 2C$ .........(4)

The reaction (1) is multiplied with 2 and added to the reaction (2). Then reaction (3) is subtracted to obtain reaction (4).

Hence, the expression for the enthalpy change for the reaction (4) is as given below.

$\Delta H_{(4)} = 2\Delta H_{(1)}+\Delta H_{(2)}-\Delta H_{(3)}=300 - 125 - 350 = - 175$

Hence, for

$B + D \rightarrow E + 2C, \Delta H$ will be

$-175\ {kJ}/{mol}$ .

One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure, The change in internal energy of the gas during the transition is:

0%
$20\ J$
0%
$-12\ kJ$
0%
$20\ kJ$
0%
$-20\ kJ$

Explanation

$\Delta U= nC_vdT$ whene

$n$ is no. of moles,

$C_v$ is heat capacity at constant volume and

$dT$ is temperature change
for diatomic gas

$C_v=\dfrac{3}{2}R$
from ideal gas equation

$T_1=\dfrac{P_1V_1}{R}$ and

$T_2=\dfrac{P_2V_2}{R}$

$\Delta U= C_v(T_2-T_1)$

$\Delta U= \dfrac{3}{2}R(\dfrac{P_2V_2}{R}-\dfrac{P_1V_1}{R})$

$\Delta U=\dfrac{3}{2}(P_2V_2-P_1V_1)= -12\ kJ$
hence correct answer is option B

For the reaction,

$X_2O_4(i)\rightarrow 2XO_2(g)$ ,

$\Delta U=2.1k cal, \Delta s=20 cal K^{-1}$ at 300 K.

Hence,

$\Delta G$ is:

0%
2.7 k cal
0%
-2.7 k cal
0%
9.3 k cal
0%
-9.3 kcal

Explanation

The relationship between the enthalpy change and the change in internal energy is as shown below.

$\Delta H=\Delta U+\Delta n_gRT$

Substitute values in th above expression.

$=2.1+\dfrac {2\times 2\times 300}{1000}=3.3$

The relationship between the change in free energy, the enthalpy change and the entropy change is as shown below.

$\Delta G=\Delta H-T\Delta s$

$=3.3-300\times \dfrac {26}{1000}=3.3-6=-2.7 K cal$

Hence, the change in the free energy is -2.7 K cal.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Explanation

$\Delta G=\Delta H-T\Delta S$

Decrease in randomness,

$\Delta S = -ve$

$-ve =\Delta H-[T\times (-ve)]$ (as spontaneous)

Exothermic reactions may be accompanied by an increase or decrease of randomness.

The thermal motion means

0%
motion due to heat engine
0%
disorderly motion of the body as a whole
0%
motion of the body that generates heat
0%
random motion of molecules

Assertion (A) : Zeroth law of thermodynamics gives us the concept of energy
Reason (R) : Internal energy is dependent on temperature

0%
1) Both Assertion and Reason are true and reason is correct explanation of Assertion
0%
2) Both Assertion and Reason are true but reason is not correct explanation of Assertion
0%
3) Assertion is true but reason is false
0%
4) Assertion is false but reason is true

Internal energy per mole of gas depends on

0%
viscosity
0%
density
0%
temperature
0%
thermal conductivity

The internal energy of an isolated system :

0%
remains constant
0%
keeps on changing
0%
zero
0%
may change depending on gas

Explanation

From 1st law of thermodynamics,

$\Delta Q = \Delta U + \Delta W$

As the system is isolated from the surroundings, there is no heat transfer and the volume will remain constant.

Thus, work done is zero.

Using

$First \ Law \ of \ Thermodynamics$ , internal energy will be constant.

The laws of thermodynamics speak about

0%
rates of chemical changes
0%
feasibility and energy transformations of a

process
0%
Both the rate and energy changes of a process
0%
Energy changes in chemical reactions only

The internal energy of a perfect gas depends on :

0%
Pressure
0%
Temperature
0%
Volume
0%
Specific heat

Explanation

According to the assumptions of $Kinetic \ Theory \ of \ Gases$ , the molecules do not occupy any volume and attractive forces are absent between molecules of an ideal gas.

Thus, other parameters such as volume, pressure etc do not play a significant role in the internal energy of an ideal gas and is a function of temperature only.

For an ideal gas,

$U=U(T)$

where, U is the internal energy of the ideal gas.

T is the temperature of the ideal gas.

The internal energy of an ideal gas depends upon

0%
only its pressure
0%
only its volume
0%
only its temperature
0%
its pressure and volume

Explanation

$\textbf{Explanation -}$

Since, we know that - internal energy is sum of a gas of translation

kinetic energy and rotational kinetic energy and vibration energy.

But as a given - for ideal gas -

the gas possesses only translational kinetic energy

$\bullet$ Since we know that - energy of a gas per mole

per degree of freedom

$= \dfrac {1}{2} RT$

So,

$U = \text {internal energy} = kE_{x} + kE_{y} + kE_{z}$

$U = \dfrac {1}{2} RT + \dfrac {1}{2}RT + \dfrac {1}{2}RT$

$U = \dfrac {3}{2} RT \rightarrow U\propto T$

Hence option (C) is correct.

Internal energy does not include

0%
vibrational energy
0%
rotational energy
0%
energy arising by gravitational pull
0%
nuclear energy

Which of the following is not a thermodynamic coordinate?

0%
Gas constant (R)
0%
Pressure (P)
0%
Volume (V)
0%
Temperature (T)

Gas constant

$(R)$ equals to

0%
$\dfrac{C_p}{C_v}$
0%
$1$
0%
$C_v-C_p$
0%
$C_p-C_v$

Explanation

By Mayer's formula , we have

$C_{P}-C_{V}=R$

Where

$C_{P}$ and

$C_{V}$ are the molar specific heats of gas at constant pressure and constant volume respectively .

$C_v$ for ammonia gas is

$(in J \ mol^{-1} \ K^{-1})$

0%
$29.0$
0%
$12.5$
0%
$20.7$
0%
$31.2$

Explanation

Heat capacity of ammonia at constant volume is

$29J mol^{-1}K^{-1}$

The heat change in a chemical reaction at constant volume is given by:

0%
$\Delta H$
0%
$\Delta E$
0%
$\Delta T$
0%
$\Delta V$

Explanation

First Law of Thermodynamics dictates that

ΔQ=ΔE+ΔWΔQ=ΔU+ΔW

At constant V,

ΔW=0ΔW=0

So,

$\Delta Q = \Delta E$

At the boiling point of water the saturated vapour pressure will be (in mm of Hg)

0%
750
0%
760
0%
850
0%
860

$10\ g$ of liquid at

$300\ K$ is heated to

$350\ K$ . The liquid absorbs

$6\ kcals$ . What is the specific heat of the liquid (in

$cal/g\ ^o C$ )?

0%
$6$
0%
$12$
0%
$60$
0%
$120$

Explanation

From the equation of Specifc heat we know,

$C_p$

$=$

$Q$

$/$ (

$m$

$\Delta T$ )

Where,

$C_p$ is the specific heat of the liquid

$m$ is the mass of the liquid that is

$10grams$

$\Delta T$ is the change in temperature from

$300K$ to

$350K$ that is

$50K$ or

$50^oC$ , because per degree change in Kelvin is the same as per degree change in Celcius scale.

$Q$ is the amount of heat absorbed, which is

$6$ kcals or

$6000$ cals.

Putting all the values in the equation we get,

$C_p$

$=$

$6000$

$/$ (

$10$

$*$

$50$ )

This comes out to be equal to

$12$

$cal$

$/$

$g^oC$

Option B is the correct answer.

Joule's experiment converts

0%
work into heat
0%
work into electricity
0%
heat into work
0%
electricity into work

What describes a spontaneous reaction?

0%
Positive $\Delta H$
0%
Negative $\Delta H$
0%
Positive $\Delta G$
0%
Negative $\Delta G$

Explanation

Option D is the correct answer.

For a reaction to be spontaneous, the value of

$\Delta G$ should be Negative.

The temperature at the bottom of a high water fall is higher than that at the top because

0%
by itself heat flows from higher to lower temperature
0%
the difference in height causes a difference in pressure
0%
thermal energy is transformed into mechanical energy
0%
mechanical energy is transformed into thermal energy

Therm is the unit of

0%
heat
0%
temperature
0%
thermometry
0%
work

Combustion is

0%
exothermic reaction
0%
endothermic reaction
0%
addition reaction
0%
None of these

Explanation

$\textbf{Explanation:}$

$\bullet$ Combustion is an oxidation reaction and during this reaction, heat is produced. In this type of reaction, bonds are broken first and then new bonds may be created results in new material formation. So, it is always an exothermic reaction.

$Answer:$

Hence, option A is the correct option.

Which one of the following is not a thermodynamical co-ordinate?

0%
$V$
0%
$R$
0%
$T$
0%
$P$

Explanation

$R$ is a constant term. To define a thermo dynamic state of a gas we use any two of three physical quantities

$P,V$ and

$T$ . Following ratio is always constant

$\cfrac{PV}{T}=R$ (constant)
So, if we change

$P$ and

$V, T$ will automatically change itself to make ratio constant.

Fill in the blank.
Thermodynamics is the branch of science concerned with ____ and ________ and their relation to energy and work.

0%
heat, temperature
0%
temperature, pressure
0%
heat, volume
0%
volume, pressure

Identify the incorrect statement ?

0%
Energy can be converted into matter
0%
Matter can be converted into energy
0%
Energy can be converted into matter, and matter can be converted into energy
0%
Matter can be converted into energy, but energy cannot be converted into matter
0%
Energy can be measured in units of joules and calories

Explanation

Energy and matter are interconvertible and the relation between them is given by

$E=mc^2$ as followed from Einstein's theory of Relativity.

The conversion of energy, kinetic or electromagnetic, to particles with rest mass is experimentally observed only when the energy comes in big enough clumps to create a massive particle, when the energy is low, this is going to be the lightest charged particle, the electron. Electrons can only be created along with a positively charged particle, to conserve charge, and this is almost always a positron.

The conversion of mass to energy is nowadays popular in nuclear fission reactors.

If there were no atmosphere, the average temperature on earth surface would be

0%
lower
0%
higher
0%
same
0%
${{0}^{o}}C$

Mixture of ice and water is form a

0%
closed system
0%
open system
0%
isolated system
0%
heterogeneous system

Free energy change for the process

$A(s)\rightleftharpoons B(l)$ will be:

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

$\Delta U^{ \ominus }$ of combustion of methane is

$-X kJ mol^{-1}$ The value of

$\Delta H^{ \ominus }$ is

0%
$= \Delta U^{ \ominus }$
0%
$> \Delta U^{ \ominus }$
0%
$< \Delta U^{ \ominus }$
0%
$0$

Explanation

The balanced chemical equation for the combustion reaction is

$CH_4(g) +2O_2(g) \longrightarrow CO_2(g)+2H_2O(l)$

$\Delta_{ng} =1-3= -2.$

$\Delta H^{ \ominus } =\Delta U^{ \ominus }+ \Delta _{ng} RT= \Delta U^{ \ominus }-2RT$

$\therefore \Delta H^{ \ominus } < \Delta U^{ \ominus }$ OR (iii) is the correct answer

The internal energy of compressed real gas, as compared to that of the ideal gas at the same temperature is

0%
less
0%
more
0%
sometimes less, sometimes more
0%
maximum

The internal energy of a perfect gas is :

0%
completely kinetic
0%
completely potential
0%
sum of potential and kinetic energy of the molecules
0%
difference of kinetic and potential energy of the molecules

Which of the following process is non-spontaneous?

0%
Heat flow from hot end to cool end.
0%
Water flow from higher level to lower level.
0%
Gas flow from lower pressure region to higher pressure region.
0%
Gas flow from higher pressure region to lower pressure region.

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 1 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 1 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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