Under which of the following conditions is the relation, $$\Delta H=\Delta E+P\Delta V$$ valid for a system:

Constant temperature and pressure

Constant temperature, pressure and composition

For the reaction, $$2A(g)+B(g)\rightarrow 2D(g),\Delta {U}^{o}=-10.5kJ$$ and $$\Delta {S}^{o}=-44.10J{K}^{-1}$$. Calculate $$\Delta {G}^{o}$$ for the reaction and predict whether the reaction may occur spontaneously:

$$0.165\ kJ$$, the reaction is not spontaneous.

$$0.225\ kJ$$, the reaction is not spontaneous.

$$0.164\ kJ$$, the reaction is spontaneous.

$$0.225\ kJ$$, the reaction is spontaneous.

Assertion : In Free expansion, $$\Delta U = 0$$ Reason : No work is done in free expansion

If both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

If both assertion and reason are false.

$$5$$ mole of ideal gas at $$100K$$ ($${C}_{v,m}=28J/mol/K$$). It is heated upto $$200K$$. Calculate $$\Delta U$$ and $$\Delta (PV)$$ for the process. ($$R=8J/mol-K$$)

$$\Delta U=14kJ;\Delta (PV)=4KJ$$

$$\Delta U=28kJ;\Delta (PV)=8KJ$$

$$\Delta U=14kJ;\Delta (PV)=8KJ$$

$$\Delta U=28kJ;\Delta (PV)=4KJ$$

The Normal boiling point of a liquid 'X' is $$400$$ K. Which of the following statement is true about the process $$X(l) \rightarrow X(g)$$?

At $$400$$ K and 1 atm pressure $$\Delta G=0$$

At $$400$$K and 2 atm pressure $$\Delta G=+ve$$

At $$400$$K and 0.1 atm pressure $$\Delta G=-'ve$$

At $$410$$K and 1 atm pressure $$\Delta G=+ve$$

For hydrogenation of ethene into ethane as per following chemical reaction $$CH_2=CH_2+H_2 \xrightarrow[]{Pd} CH_3-CH_3$$ The $$\Delta H_{reaction}$$ is written as Calculate $$\Delta H_{reaction}$$ for

$$\Delta H_{reaction}=4 \Delta H_{C-H}+ \Delta H_{C=C} + \Delta H_{H-H}-6 \Delta H_{H-H}-2 \Delta H_{C-C}$$

$$\Delta H_{reaction}=4 \Delta H_{C-H}+ 2\Delta H_{C-C} + \Delta H_{H-H}-6 \Delta H_{C-H}- \Delta H_{C-C}$$

$$\Delta H_{reaction}=4 \Delta H_{C-H}+ \Delta H_{C=C} + \Delta H_{H-H}-6 \Delta H_{C-H}- \Delta H_{C-C}$$

$$\Delta H_{reaction}= \Delta H_{C-H}+ \Delta H_{C=C} + \Delta H_{H-H}- \Delta H_{C-H}- \Delta H_{C-C}$$

For which of the following reaction $$\Delta H$$ is equal to $$\Delta u$$?

$$ 2HI(g) \leftrightharpoons H_2(g)+I_2(g)$$

$$ 2NO_2(g) \leftrightharpoons N_2O_4(g)$$

$$N_2(g) + 3H_2(g) \leftrightharpoons 2 NH_3(g)$$

$$ 2SO_2(g) +O_2(g) \leftrightharpoons 2SO_3(g)$$

A gas is expanded from volume $$ V_0$$ and $$2V_0 $$ under three different processes. Process 1 is isobaric, process 2 is isothermal and process 3 is adiabatic . Let $$ \triangle U_1, \triangle U_2 $$ and $$\triangle U_3 $$ be the change in internal energy of the gas in these process. then :

$$ \triangle U_1 > \triangle U_2 > \triangle U_3 $$

$$ \triangle U_1 < \triangle U_2 < \triangle U_3 $$

$$ \triangle U_2 < \triangle U_1 < \triangle U_3 $$

$$ \triangle U_2 < \triangle U_3 < \triangle U_1 $$

MCQ Questions for Class 11 Medical Chemistry Thermodynamics Quiz 8

2 H_{2 (g)} + O_{2 (g)} \to 2 H_{2} O_{(l)}; Δ H = - v e

$2{H}_{2(g)}+{O}_{2(g)}\rightarrow 2{H}_{2}{O}_{(l)};\Delta H=-ve$ and

Δ G = - v e

$\Delta G=-ve$ . Then the reaction is

0%
Spontaneous and instantaneous
0%
Spontaneous and endothermic
0%
Spontaneous and slow
0%
Non spontaneous and slow

Answer the question based on the diagram involving

1

$1$ mole of ideal gas.
Process

A \to B

$A\rightarrow B$ represents:

0%
isobaric
0%
isochoric
0%
isothermal
0%
adiabatic

Explanation

Solution:- (B) isochoric process

In process

A \to B

$A \rightarrow B$ , as the volume is constant, thus the process represents isochoric process.

In standard state the non spontaneous reaction among the following is:

0%
melting of ice
0%
natural radioactivity
0%
freezing of water
0%
rusting of iron

Explanation

Ans

C

$C$ - freeging of water

water exists in liquid state naturally at standard state

25^{\circ} c = 298 k

${ 25 }^{ \circ }c=298k$ . Hence freezing of

H_{2} O

${ H }_{ 2 }O$ at 298k is non-spontaneous as ice will absorb heat from surroundings thereby gets converted to the liquid state as it is a natural process of is irreversible.

Under which of the following conditions is the relation,

Δ H = Δ E + P Δ V

$\Delta H=\Delta E+P\Delta V$ valid for a system:

0%
Constant pressure
0%
Constant temperature
0%
Constant temperature and pressure
0%
Constant temperature, pressure and composition

The heat capacity of 2.0 moles of He gas at constant temperature is (in cal/mol/K)

0%
0
0%
3
0%
8
0%
$\infty$

For the reaction,

2 A (g) + B (g) \to 2 D (g), Δ U^{o} = - 10.5 k J

$2A(g)+B(g)\rightarrow 2D(g),\Delta {U}^{o}=-10.5kJ$ and

Δ S^{o} = - 44.10 J K^{- 1}

$\Delta {S}^{o}=-44.10J{K}^{-1}$ . Calculate

Δ G^{o}

$\Delta {G}^{o}$ for the reaction and predict whether the reaction may occur spontaneously:

0%
$0.165\ kJ$ , the reaction is not spontaneous.
0%
$0.225\ kJ$ , the reaction is not spontaneous.
0%
$0.164\ kJ$ , the reaction is spontaneous.
0%
$0.225\ kJ$ , the reaction is spontaneous.

Evaporation of water is

0%
An endothermic change
0%
An exothermic change
0%
A process where no heat change occurs
0%
A process accompained by chemical reaction

27 C

$27C$ the reaction,

C_{6} {H_{6}}_{(l)} + \frac{15}{2} {O_{2}}_{(g)} \to {6 C O}_{2}_{(g)} + {{3 H}_{2} O}_{(l)}

${ C }_{ 6 }{ { H }_{ 6 } }_{ { (l) } }+\frac { 15 }{ 2 } { { O }_{ 2 } }_{ (g) }\rightarrow { { 6CO }_{ 2 } }_{ (g) }+{ { 3H }_{ 2 }O }_{ (l) }$
proceeds spontaneously because the magnitude of?

0%
$\Delta H=T\Delta S$
0%
$\Delta H>T\Delta S$
0%
$\Delta H$ < $T\Delta S$
0%
None of these

Explanation

Δ H < T Δ S

$\Delta{H} < T \Delta{S}$

As we know that,

Δ G = Δ H - T Δ S

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

As the reaction is proceeding spontaneously,

Δ G

$\Delta{G}$ must be negative.

Therefore, for

Δ G

$\Delta{G}$ to be negative,

Δ H < T Δ S

$\Delta{H} < T \Delta{S}$

For the given reaction

H_{2} (g) + S (s) ⟶ H_{2} S (g); Δ H_{T} = 100 k J / m o l

$H_{2}(g)+S(s)\longrightarrow H_{2}S(g);\ \Delta H_{T}=100\ kJ/mol$ and

Δ S_{T} = 400 J / m o l / K

$\Delta S_{T}=400\ J/mol/K$
Temperature at which above reaction occurs reversibly is (Assuming

Δ H_{T}

$\Delta H_{T}$ and

Δ S_{T}

$\Delta S_{T}$ are independent of temperature)

0%
$200\ K$
0%
$250\ K$
0%
$400\ K$
0%
$None$

Explanation

We know that

Δ S_{T} = \frac{Δ H_{T}}{T}

$\Delta S_T=\cfrac {\Delta H_T}{T}$

Given,

Δ S_{T} = 400 J m o l^{- 1} K^{- 1}

$\Delta S_T= 400Jmol^{-1}K^{-1}$

Δ H_{T} = 100 K J m o l^{- 1}

$\Delta H_T= 100 KJmol^{-1}$

T = \frac{Δ H_{T}}{Δ S_{T}} = \frac{100 \times 1000}{400} = 250 K

$T= \cfrac {\Delta H_T}{\Delta S_T}= \cfrac {100 \times 1000}{400}=250K$

Hence, Temperature is

250 K

$250K$ .

A constant volume gas thermometer shows pressure reading of 50 cm and 90 cm of mercury at 0

^{0} C

$^{0}C$ and 100

^{0} C

$^{0}C$ respectively. When the pressure reading is 60 cm of mercury, the temperature is:-

0%
25 $^{0}C$
0%
40 $^{0}C$
0%
$^{0}C$
0%
$30^{0}C$

Explanation

Given,

Temperature ${{T}_{1}}={{0}^{o}}C$ and height ${{h}_{1}}=50\,cm$

Temperature ${{T}_{2}}=100{{\,}^{o}}C$ and height ${{h}_{2}}=90\,cm$

$\dfrac{T-{{T}_{1}}}{h-{{h}_{1}}}=\dfrac{{{T}_{2}}-{{T}_{1}}}{{{h}_{2}}-{{h}_{1}}}$

$\dfrac{T-0}{60-50}=\dfrac{100-0}{90-50}$

$T=25{{\,}^{o}}C$

Hence, at $60 cm$ , temperature is $25{{\,}^{o}}C$

A gas occupies

2 L

$2L$ at S.T.P. It is provided with

300 J o u l e

$300 Joule$ of heat so that its volume becomes

2.5 l i t r e

$2.5 litre$ at a pressure of

1 a t m

$1 atm$ . The value of

Δ U

$\Delta U$ (Change in internal energy) of process is :-

0%
$350.5 Joule$
0%
$249.5 Joule$
0%
$150.35 Joule$
0%
None of these

Explanation

Given:

Initial volume of gas,

V_{i} = 2 l

$V_i =2 l$

Heat provided, Q= 300J

Final volume,

V_{f} = 2.5 l

$V_f = 2.5 l$

Final pressure,

P_{f} = 1 a t m

$P_f =1 atm$

Solution:

We know that the work is done at constant pressure and thus is irreversible.

Work done, W=-P(Change in volume)

\Rightarrow W = - 1 \times (2.5 - 2)

$\Rightarrow W=-1×(2.5-2)$

\Rightarrow W = - 0.5 a t m - l

$\Rightarrow W= -0.5 atm-l$

\Rightarrow W = - 0.5 \times 101325 \times \frac{1}{1000} J

$\Rightarrow W= -0.5 ×101325 ×\dfrac{1}{1000} J$

\Rightarrow W = - 50.6625 J

$\Rightarrow W=-50.6625 J$

From the first law of thermodynamics,

Q = δ U - W

$Q=\delta U -W$ , where

δ U = C h a n g e i n i n t e r n a l e n e r g y

$\delta U = Change in internal energy$

\Rightarrow δ U = Q + W

$\Rightarrow \delta U = Q+ W$

\Rightarrow δ U = 300 + (- 50.6625)

$\Rightarrow \delta U = 300 +(-50.6625)$

Therfore, $$\delta U = 249.3Joule$$

Hence, correct option is B.

The values of

Δ H

$\Delta H$ and

Δ S

$\Delta S$ for the reaction,

C_{(g r a p h i t e)} + {C O}_{2_{(g)}} ⟶ {2 C O}_{(g)}

${ C }_{ \left( graphite \right) }+{ CO }_{ { 2 }_{ \left( g \right) } }\longrightarrow { 2CO }_{ \left( g \right) }$ are

170 k J

$170 \ kJ$ and

170 J K^{- 1}

$170 \ JK^{-1}$ , respectively. This reaction will be spontaneous at :-

0%
$510 \ K$
0%
$710 \ K$
0%
$910 \ K$
0%
$1110 \ K$

Among the following the non-spontaneous reaction is :

0%
$ROH+R'MgX \rightarrow R'H+Mg(OR)X$
0%
$RONa+NH_{3} \rightarrow NaHN_{2}+ROH$
0%
$RONa+H_{2}O \rightarrow NaOH+ROH$
0%
$ROH+HC \equiv CNa \rightarrow RONa+HC \equiv CH$

In a cyclic process, the change in the internal energy of a system over one complete cycle

0%
depends on the path
0%
is always negative
0%
is always zero
0%
is always positive

A coffee cup calorimeter initially contains

125 g

$125\ g$ of water, at a temperature of

{24.2}^{o} C

$24.2^oC$ .

8 g

$8\ g$ of ammonium nitrate

(N H_{4} N O_{3})

$(NH_4NO_3)$ , also at

{24.2}^{o} C

$24.2^oC$ , is added to the water and the final temperature is

{18.2}^{o} C

$18.2^oC$ . What is the heat of solution of ammonium nitrate in

k J / m o l

$kJ/mol$ ? The specific in capacity of the solution is

4.2 J /^{o} C g

$4.2\ J/^oCg$ .

0%
$33.51\ kJ/mol$
0%
$39.5\ kJ/mol$
0%
$33.2\ kJ/mol$
0%
$37.3\ kJ/mol$

If the temperature difference

T_{1} - T_{2}

$T_1 - T_2$ is

120^{o} C

$120^oC$ . the temperature difference between points A & B is

0%
30
0%
45
0%
60
0%
75

When 250 KJ heat is the system then it does work of 400 KJ then calculate change in internal energy of system ?

0%
650 KJ
0%
-650 KJ
0%
200 KJ
0%
-200 KJ

1 g of a steam at

100^{_{-}^{0}} C

$100^{_-^0}C$ melt how much ice at

0^{_{-}^{0}} C

$0^{_-^0}C$ ? (Latent heat of ice = 80 cal/gm and latent heat of steam = 540 cal/gm)-

0%
1 gm
0%
2 gm
0%
4 gm
0%
8 gm

Explanation

Suppose

m

$m$ gm ice melted,

then heat required for its melting =

m L = m \times 80 c a l

$mL=m×80cal$ .

heat avaiable with steam for being condensed and then brought to

0^{\circ} C .

$0^∘C.$

= 1 \times 540 + 1 \times 1 \times (100 - 0) = 640 c a l

$=1×540+1×1×(100-0)=640cal$

⇒ heat lost = heat taken

⇒

640 = m \times 80 \Rightarrow m = 8 g m

$640=m×80⇒m=8gm$ .

One gram of ice (at

0^{o} C

$0^{o}C$ ) is mixed with one gram of steam (at

100^{o} C

$100^{o}C$ ). After thermal equilibrium, the temperature of the mixture is

0%
$0^{o}C$
0%
$100^{o}C$
0%
$55^{o}C$
0%
$80^{o}C$

Explanation

Total steam does not condense

∴

$\therefore$ Final temperature

= 100^{0} C

$={ 100 }^{ 0 }C$

Which of the following can be calculated from Born-Haber cycle for

A l_{2} O_{3}

$Al_2O_3$ ?

0%
Lattice energy of $Al_2O_3$
0%
Electron affinity of O-atom
0%
Ionisation energy of Al
0%
All of these

Internal energy of an ideal gas depends on:

0%
pressure
0%
temperature
0%
volume
0%
none of these

Explanation

The relation between the internal energy and enthalpy, it can be derived that the internal energy of the gas is independent of the volume and pressure whereas it is only temperature-dependent.

Option

B

$B$ is correct.

What is the change in entropy when

2.5

$2.5$ mole of water is heated from

27^{o} C

$27^{o}C$ to

87^{o} C

$87^{o}C$ ? Assume that the heat capacity is constant

(C_{p . m} (H_{2} O) = 4.2 J / g - K I n (1.2) = 0.18)

$(C_{p.m}(H_{2}O)=4.2\ J/g-K\ In (1.2)=0.18)$

0%
$16.6\ J/K$
0%
$9\ J/K$
0%
$34.02\ J/K$
0%
$1.89\ J/K$

Calculate

Δ_{r} G

${ \Delta }_{ r }G$ for the reaction at

2 \overset{˚}{7} C

$2\mathring { 7\quad } C$ :-

H_{2 (g)} + 2 A g_{(a q)}^{+} ⇌ 2 A g_{s} + 2 H_{(a q)}^{+}

${ H }_{ 2(g) }+2A{ g }_{ (aq) }^{ + }\rightleftharpoons 2A{ g }_{ s }+2{ H }_{ (aq)}^{+}$
Given :

P_{H_{2}} = 0.5

${ P }_{ { H }_{ 2 } }=0.5$ bar;

[A g^{+}] = 1 0^{- 5} M

${ [A{ g }^{ + }] }=1{ 0 }^{ -5 }M$

[H^{+}] = 1 0^{- 3} M

${ [{ H }^{ + }] }=1{ 0 }^{ -3 }M$ ;

△_{r} \overset{˚}{G} [A g^{+}] = 77.1 K J / m o l

${ \triangle }_{ r }\mathring { G\quad \quad } [A{ g }^{ + }]=77.1\quad KJ/mol$

0%
-154.2 KJ/mol
0%
-178.9 KJ/mol
0%
-129.5 KJ/mol
0%
none of these

Δ G

$\Delta G$ for a reaction is negative, the change is:

0%
reversible
0%
spontaneous
0%
equilibrium
0%
non-spontaneous

Explanation

Solution:- (B) Spontaneous

For a spontaneous reaction,

Δ G = - v e

$\Delta{G} = -ve$

A vessel contains

N

$N$ molecules of a gas at temperature

T

$T$ . Now the number of molecules is doubled, keeping the total energy in the vessel constant. The temperature of the gas is.

0%
$T$
0%
$2T$
0%
$\dfrac{T}{2}$
0%
$\sqrt 2 T$

Explanation

Total energy

= n k T

$=nkT$

n^{'} = 2 n

$n'=2n$ (Since the 'P' is in closed box)

T^{'} = \frac{T}{2}

$T'=\cfrac { T }{ 2 }$

P V = n R T

$PV=nRT$

∴ T^{'} = \frac{T}{2}

$\therefore T'=\cfrac { T }{ 2 }$

Determine

Δ U^{o}

$\Delta U^o$ at 300 K for the following reaction using the listed enthalpies of reaction :

4 C O (g) + 8 H_{2} (g) \to 3 C H_{4} (g) + C O_{2} (g) + 2 H_{2} O (I)

$4CO(g)+8H_{ 2 }(g)\rightarrow 3CH_{ 4 }(g)+CO_{ 2 }(g)+2H_{ 2 }O(I)$

C (g r a p h i t e) + 1 / 2 O_{2} (g) \to C O (g); Δ H_{1}^{o} = - 110.5 k J

$C (graphite)+1/2O_2(g) \rightarrow CO(g); \Delta H^o_1 = -110.5 kJ$

C O (g) + 1 / 2 O_{2} (g) \to C O_{2} (g); Δ H_{2}^{o} = - 282.9 k J

$CO(g) +1/2 O_2(g) \rightarrow CO_2(g); \Delta H^o_2 = -282.9 kJ$

H_{2} (g) + 1 / 2 O_{2} (g) \to H_{2} O (I); Δ H_{3}^{o} = - 285.8 k J

$H_2(g) + 1/2 O_2(g) \rightarrow H_2O(I); \Delta H^o_3 = -285.8 kJ$

C (g r a p h i t e) + 2 H_{2} (g) \to C H_{4} (g); Δ H_{4}^{o} = - 74.8 k J

$C(graphite)+2H_2(g) \rightarrow CH_4(g); \Delta H^o_4 = -74.8 kJ$

0%
$-653.5 kJ$
0%
$-686.2kJ$
0%
$-747.4kJ$
0%
None of these

For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter

Δ U

$\Delta U$ and w correspond to

0%
$\Delta U<0,w=0$
0%
$\Delta U<0,w<0$
0%
$\Delta U>0,w=0$
0%
$\Delta U>0,w>0$

An ideal gas is taken through a cyclic thermo dynamical process through four steps. The amounts of heat involved in steps are

Q_{1} = 5960 J, Q_{2} = - 5585 J, Q_{3} = - 2980 J, Q_{4} = 3645 J

$Q_1 = 5960 J, Q_2 = -5585 J, Q_3 = -2980 J, Q_4 = 3645 J$ ; respectively, the corresponding works involved are

W_{1} = 2200, W_{2} = - 825 J, W_{3} = - 1100 J

$W_1 = 2200 , W_2 = -825 J, W_3 = -1100 J$ and

W_{4}

$W_4$ respectively. Find the value of

W_{4}

$W_4$ and efficiency of the cycle

0%
1315 J, 10%
0%
765 J, 11%
0%
both of them
0%
none of these

Explanation

Given;-

\begin{matrix} Q_{1} = 5960 J, W_{1} = 2200 J \\ Q_{2} = 5585 J, W_{2} = 825 J \\ Q_{3} = 2980 J, W_{3} = 1100 J \\ Q_{4} = 3645 J, W_{4} = f i n d \\ η = f i n d \end{matrix}

$\begin{matrix} { Q_{ 1 } }=5960J\, \, \, ,{ W_{ 1 } }=2200J \\ { Q_{ 2 } }=5585J\, \, \, ,{ W_{ 2 } }=825J \\ { Q_{ 3 } }=2980J\, \, \, ,{ W_{ 3 } }=1100J \\ { Q_{ 4 } }=3645J\, \, \, ,{ W_{ 4 } }=find \\ \eta =find \\ \end{matrix}$

In cycic process

Δ U = 0

$\Delta U=0$

A / q

$A/q$

1 s t

$1st$ law of thermodynamics

Δ Q = Δ U + W

$\Delta Q= \Delta U+W$

\Rightarrow + 1040 = 275 + W_{4}

$\Rightarrow +1040 = 275+W_4$

\Rightarrow W_{4} = 1040 - 275

$\Rightarrow W_4=1040-275$

= 765

$=765$

η % = \frac{o u t p u t}{i n p u t} \times 100

$\eta \% = \dfrac{{output}}{{input}} \times 100$

= \frac{1040}{5960 + 3645}

$= \dfrac{{1040}}{{5960 + 3645}}$

= \frac{208}{1321} \times 100

$= \dfrac{{208}}{{1321}} \times 100$

Hence, we get

η = 10.82 %

$\eta = 10.82\%$

Assertion : In Free expansion,

Δ U = 0

$\Delta U = 0$
Reason : No work is done in free expansion

0%
If both assertion and reason are true and reason is the correct explanation of assertion.
0%
If both assertion and reason are true but reason is not the correct explanation of assertion.
0%
If assertion is true but reason is false.
0%
If both assertion and reason are false.

For reversible vaporisation of water at

10 0^{o} C

${\text{10}}{{\text{0}}^{\text{o}}}{\text{C}}$ and 1 atmospheric pressure,

Δ G

$\Delta {\text{G}}\;$ is equal

0%
$\Delta {\text{H}}\;$
0%
$\Delta {\text{S}}\;$
0%
zero
0%
$\Delta {\text{H/T}}\;\;$

5

$5$ mole of ideal gas at

100 K

$100K$ (

C_{v, m} = 28 J / m o l / K

${C}_{v,m}=28J/mol/K$ ). It is heated upto

200 K

$200K$ . Calculate

Δ U

$\Delta U$ and

Δ (P V)

$\Delta (PV)$ for the process. (

R = 8 J / m o l - K

$R=8J/mol-K$ )

0%
$\Delta U=28kJ;\Delta (PV)=8KJ$
0%
$\Delta U=14kJ;\Delta (PV)=4KJ$
0%
$\Delta U=14kJ;\Delta (PV)=8KJ$
0%
$\Delta U=28kJ;\Delta (PV)=4KJ$

Explanation

Δ U = n C_{v} Δ T = \frac{5 \times 28 \times 100}{1000} = 14 K J

$\Delta U=n{ C }_{ v }\Delta T=\cfrac { 5\times 28\times 100 }{ 1000 } =14KJ$

Δ (P V) = \frac{5 \times 8 \times 100}{1000} = 4 K J

$\Delta (PV)=\cfrac { 5\times 8\times 100 }{ 1000 } =4KJ\quad$

A diatomic ideal gas is expanded at constant at pressure. If work done by the system is

10 J

$10 \,J$ then calculate heat absorbed.

0%
$40 \,J$
0%
$20 \,J$
0%
$35 \,J$
0%
$15 \,J$

Explanation

Δ U = n C v_{m} Δ T

$\Delta U = nCv_m \Delta T$

W = - n R Δ T

$W = -nR\Delta T$

- 10 = - n R Δ T

$-10 = -nR\Delta T$

∴ Δ U = n \times \frac{5}{2} R \times \frac{10}{n R}

$\therefore \Delta U = n \times \dfrac{5}{2}R \times \dfrac{10}{nR}$ or

Δ T = \frac{10}{n R}

$\Delta T = \dfrac{10}{nR}$
or

Δ U = 25

$\Delta U = 25$

∴ Δ U = q + W

$\therefore \Delta U = q + W$

25 = q - 10 \Rightarrow q = 35 J

$25 = q - 10 \Rightarrow q = 35 \,J$

The Normal boiling point of a liquid 'X' is

400

$400$ K. Which of the following statement is true about the process

X (l) \to X (g)

$X(l) \rightarrow X(g)$ ?

0%
At $400$ K and 1 atm pressure $\Delta G=0$
0%
At $400$ K and 2 atm pressure $\Delta G=+ve$
0%
At $400$ K and 0.1 atm pressure $\Delta G=-'ve$
0%
At $410$ K and 1 atm pressure $\Delta G=+ve$

For hydrogenation of ethene into ethane as per following chemical reaction

C H_{2} = C H_{2} + H_{2} \overset{P d}{\to} C H_{3} - C H_{3}

$CH_2=CH_2+H_2 \xrightarrow[]{Pd} CH_3-CH_3$
The

Δ H_{r e a c t i o n}

$\Delta H_{reaction}$ is written as
Calculate

Δ H_{r e a c t i o n}

$\Delta H_{reaction}$ for

0%
$\Delta H_{reaction}=4 \Delta H_{C-H}+ \Delta H_{C=C} + \Delta H_{H-H}-6 \Delta H_{H-H}-2 \Delta H_{C-C}$
0%
$\Delta H_{reaction}=4 \Delta H_{C-H}+ 2\Delta H_{C-C} + \Delta H_{H-H}-6 \Delta H_{C-H}- \Delta H_{C-C}$
0%
$\Delta H_{reaction}=4 \Delta H_{C-H}+ \Delta H_{C=C} + \Delta H_{H-H}-6 \Delta H_{C-H}- \Delta H_{C-C}$
0%
$\Delta H_{reaction}= \Delta H_{C-H}+ \Delta H_{C=C} + \Delta H_{H-H}- \Delta H_{C-H}- \Delta H_{C-C}$

In which process volume increases?

0%
AB, CD
0%
AB, BC
0%
CD, DA
0%
BC, CD

Which one of the following process is non-spontaneous?

0%
Dissolution of $CuSO_{4}$ in water
0%
Reaction of $H_{2}$ and $O_{2}$ to form water
0%
Water flowing down hill
0%
Flow of electric current from low potential to high potential

Hess' Law and bond energy data can be used to calculate the enthalpy change of a reaction. Bromoethane,

C H_{2} C H_{2} B r

$CH_2CH_2Br$ , can be made by reacting ethene with hydrogen bromide.

C H_{2} = C H_{2} + H B r \to C H_{3} C H_{2} B r

$CH_2=CH_2+HBr\rightarrow CH_3CH_2Br$
What is the enthalpy change for this reaction?

0%
$-674$ kJ $mol^{-1}$
0%
$-64$ kJ $mol^{-1}$
0%
$+186$ kJ $mol^{-1}$
0%
$+346$ kJ $mol^{-1}$

Explanation

The enthalpy change is the difference of the enthalpy of product to the reactans.

$\text{The enthalpy change is the difference of the enthalpy of product to the reactans.}$

By putting the values we get  option B is the correct option.

$\text{By putting the values we get option B is the correct option.}$

The resultant heat change in a reaction is the same whether it takes
place in one or several stages. This statement is called:

0%
Lavoisier and Laplace law
0%
Hess's law
0%
Joule's law
0%
Le-chatelier's principle

An ideal gas undergoes a cyclic process as shown in figure;

Δ u_{B C} = - 5 K J m o l^{- 1}, q_{A B} = 2 K J m o l^{- 1}

$\Delta u_{BC}=-5 KJ mol^{-1}, q_{AB}= 2 KJ mol^{-1}$

w_{A B} = - 5 K J m o l^{- 1}, W_{C A} = 3 K J m o l^{- 1}

$w_{AB}= -5 KJ mol^{-1},W_{CA}=3 KJ mol^{-1}$

Heat absorbed by the system during process

C A

$CA$ is

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

For which of the following reaction

Δ H

$\Delta H$ is equal to

Δ u

$\Delta u$ ?

0%
$2HI(g) \leftrightharpoons H_2(g)+I_2(g)$
0%
$2NO_2(g) \leftrightharpoons N_2O_4(g)$
0%
$N_2(g) + 3H_2(g) \leftrightharpoons 2 NH_3(g)$
0%
$2SO_2(g) +O_2(g) \leftrightharpoons 2SO_3(g)$

Explanation

Δ n = - 1

$\Delta n =-1$
C.

Δ n = - 2

$\Delta n=-2$
D.

Δ n = - 1

$\Delta n=-1$
Option A is correct.
1599219_1738932_ans_dfd582e8e70f4f81b511a9823d8b5514.jpeg

Which statement about reactions that produce heat is not correct?

0%
Burning magnesium produces hear energy
0%
The overall reaction is exothermic
0%
The products have more energy than the reactants
0%
The temperature of the surroundings increases

Explanation

In a exothermic reaction heat is released for stability of product with less energy.

$\text{In a exothermic reaction heat is released for stability of product with less energy.}$

In which case is a reaction possible at any temperature?

0%
$\Delta H < 0, \Delta S > 0$
0%
$\Delta H < 0, \Delta S > 0$
0%
$\Delta H < 0, \Delta S > 0$
0%
None of these

Explanation

Δ G = - v e

$\Delta G=-ve$ if

Δ H = - v e

$\Delta H =-ve$ and

Δ S = + v e

$\Delta S=+ve$ .

At any temp, because of t(K)>0 always.

Hence, option A is correct.

Which of the following is the correct equation?

0%
$\Delta U=\Delta Q-W$
0%
$\Delta W=\Delta U+\Delta Q$
0%
$\Delta U=\Delta W+\Delta Q$
0%
None of these

Explanation

Δ U = Δ W + Δ Q

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

C is correct

A gas is expanded from volume

V_{0}

$V_0$ and

2 V_{0}

$2V_0$ under three different processes. Process 1 is isobaric, process 2 is isothermal and process 3 is adiabatic . Let

△ U_{1}, △ U_{2}

$\triangle U_1, \triangle U_2$ and

△ U_{3}

$\triangle U_3$ be the change in internal energy of the gas in these process. then :

0%
$\triangle U_1 > \triangle U_2 > \triangle U_3$
0%
$\triangle U_1 < \triangle U_2 < \triangle U_3$
0%
$\triangle U_2 < \triangle U_1 < \triangle U_3$
0%
$\triangle U_2 < \triangle U_3 < \triangle U_1$

Explanation

Since

U_{1}

$U_1$ is increasing in the graph than

U_{2}

$U_2$ where slope is decreasing and then

U_{3}

$U_3$ slope is more negative than

U_{2}

$U_2$ .

In a reaction

Δ H

$\Delta H$ and

Δ S

$\Delta S$ both are more than zero. In which of the following cases would the reaction be spontaneous?

0%
$\Delta H > T\Delta S$
0%
$T\Delta S > \Delta H$
0%
$\Delta H = T\Delta S$
0%
None of these

Explanation

A reaction will be spontaneous if:

Δ G < O

$\Delta G<O$

Δ H - T Δ S < 0

$\Delta H-T\Delta S < 0$

\to Δ H < T Δ S

$\rightarrow \Delta H <T\Delta S$

Which relation is correct for isochoric process

0%
$\Delta Q=\Delta U$
0%
$\Delta W=\Delta U$
0%
$\Delta Q=\Delta W$
0%
None of these

Explanation

For isochoric process

Δ V = 0 \Rightarrow Δ W = 0

$\Delta V=0\Rightarrow \Delta W=0$ From FLOT

Δ Q = Δ U + Δ W \Rightarrow Δ Q = Δ U

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

Which of the following endothermic processes are spontaneous?

0%
Melting of ice
0%
Evaporation of water
0%
Heat of combustion
0%
Both (a) and (b)

A minus sign of the free energy change denotes that:

0%
The reaction tends to proceed spontaneously
0%
The reaction is non spontaneous
0%
The system is in equilibrium
0%
The reaction is very much unlikely

The spontaneity means, having the potential to proceed without the assistance of external agency. The processes which occur spontaneously are
Note: Two or more options may be correct.

0%
flow of heat from colder to warmer body.
0%
gas in a container contracting into one comer.
0%
gas expanding to fill the available volume.
0%
burning carbon in oxygen to give carbon dioxide.

Which of the following
statements is correct?

0%
The presence of reacting species in a covered beaker is an example of open system.
0%
There is an exchange of energy as well as matter between the system and the surroundings in a closed system.
0%
The presence of reactants in a closed vessel made up of copper is an example of a closed system.
0%
The presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system.

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 8 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 8 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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