Thermodynamics Questions — JEE Chemistry

Q1

The difference between

\Delta

H and

\Delta

U (

\Delta

H –

\Delta

U), when the combustion of one mole of heptane(l) is carried out at a temperature T, is equal to :

A – 4 RT

B 3 RT

C – 3 RT

D 4 RT

Correct Answer

Option A

Solution

We know,

\Delta

H -

\Delta

U =

\Delta

ngRT C7H16(

l

) + 11O2(g) $\to$ 7CO2(g) + 8H2O(

l

) Here

\Delta

ng = 7 - 11 = - 4 $\therefore$

\Delta

H -

\Delta

U = - 4RT

Q2

5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV = 28 JK–1mol–1, calculate

\Delta

U and

\Delta

pV for this process. (R = 8.0 JK–1 mol–1]

A

\Delta

U = 14 kJ;

\Delta

(pV) = 4 kJ

B

\Delta

U = 2.8 kJ;

\Delta

(pV) = 0.8 kJ

C

\Delta

U = 14 kJ;

\Delta

(pV) = 18 kJ

D

\Delta

U = 14 kJ;

\Delta

(pV) = 0.8 kJ

Correct Answer

Option A

Solution

For ideal gas,

\Delta

U = nCv

\Delta

T = 5 $\times$ 28 $\times$ (200 - 100) = 14 kJ We know, PV = nRT $\therefore$

\Delta

(PV) = nR

\Delta

T = 5 $\times$ 8 $\times$ 100 = 4 kJ

Q3

The enthalpy change on freezing of 1 mol of water at 5oC to ice at −5oC is : (Given

\Delta

fusH = 6 kJ mol

-

1 at 0oC, Cp(H2O,

\ell

= 75.3J mol

-

1 K

-

1) Cp(H2O s) =36.8 J mol

-

1 K

-

1)

A 5.44 kJ mol

-

1

B 5.81 kJ mol

-

1

C 6.56 kJ mol

-

1

D 6.00 kJ mol

-

1

Correct Answer

Option C

Solution

(1) From 5

^\circ

C to 0

^\circ

C :

{H_1} = {C_P} \times (\Delta T)

= 75.3 \times (278 - 273)

= 377 J/mol = 0.377 KJ/mol (2) Phase change (0

^\circ

C to 0

^\circ

C) :

{H_2} = - \Delta {H_{fusion}}

= $-$ 6 KJ/mol (3) 0

^\circ

C to $-$ 5

^\circ

C (Ice phase) :

{H_3} = - {C_{P(s)}} \times \Delta T

= - 36.8 \times (273 - 268)

= $-$ 0.184 KJ/mol $\therefore$

{H_{Total}} = {H_1} + {H_2} + {H_3}

= 0.377 - 6 - 0.184

= $-$ 5.837 KJ/mol

Q4

On the basis of the following thermochemical data : (

\Delta _fG^oH^+_{(aq)}

= 0) H2O(l)

\to

H+(aq) + OH-(aq);

\Delta H

= 57.32 kJ H2(g) +

{1 \over 2} O_2(g) \to

H2O(l);

\Delta H

= -286.20 kJ The value of enthalpy of formation of OH− ion at 25oC is :

A -22.88 kJ

B -228.88 kJ

C +228.88 kJ

D -343.52 kJ

Correct Answer

Option B

Solution

Given, for reaction

(i)

\,\,\,\,\,\,{H_2}O\left( \ell \right) \to {H^ + }\left( {aq.} \right) + O{H^ - }\left( {aq.} \right);

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H_r} = 57.32\,kJ

(ii)

\,\,\,\,\,\,{H_2}\left( g \right) + {1 \over 2}{O_2}\left( g \right) \to {H_2}O\left( \ell \right);

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H_r} = - 286.20\,kJ

For reaction

(i)

\Delta {H_r} = \Delta {H^ \circ }_f\left( {{H^ + }.aq} \right) +

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H^ \circ }_f\left( {O{H^ - }.aq} \right) - \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)

57.32 = 0 + \Delta {H^ \circ }_f\left( {O{H^{ - 1}},aq} \right) -

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)\,\,\,\,\,\,...\left( {iii} \right)

For reaction

(ii)

\Delta {H_r} = \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right) - \Delta {H^ \circ }_f\left( {{H_2},g} \right) -

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{1 \over 2}\Delta {H^ \circ }_f\left( {{O_2},g} \right)

- 286.20 = \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)

On replacing this value in equ.

(iii)

we have

57.32 = \Delta {H^ \circ }_f\left( {O{H^ - }.aq} \right) - \left( { - 286.20} \right)

\Delta {H^ \circ }_f = - 286.20 + 57.32

= - 228.88\,kJ

Q5

The combustion of benzene(l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 25oC; heat of combustion (in kJ mol–1) of benzene at constant pressure will be : (R = 8.314 JK–1 mol–1)

A –3267.6

B 4152.6

C –452.46

D 3260

Correct Answer

Option A

Solution

Formula of Heat of combination is

\Delta H = \Delta u\,\, + \,\,\Delta ng\,\,RT

Where,

\Delta H

$=$ Heat of combination at constant pressure

\Delta u\, =

Heat at constant volume

\Delta {n_g}

change in number of moles for gaseous molecule. R = gas constant T = Temperature. The Required reaction C6H6(l) +

{{15} \over 2}

O2(g) $\to$ 6CO2(g) + 3H2O(l) Here O2 and CO2 are gaseous molecules so to calculate

\Delta {n_g}

we only consider those.

\therefore\,\,\,

\Delta {n_g}

$=$ 6 $-$

{{15} \over 2}

= $-$

{3 \over 2}

Given

\Delta

u = $-$ 3263.9 kJ mol $-$ 1 R = 8.314 JK $-$ mol $-$ 1 R = 8.314 JK $-$ 1 mol $-$ 1 = 8.314 $\times$ 10 $-$ 3 kJ K $-$ 1 mol $-$ 1 T = 25o C = 25 + 273 K = 298 K So,

\Delta H

= $-$ 3263.9 +

\left( { - {3 \over 2}} \right)

$\times$ 8.314 $\times$ 10 $-$ 3 $\times$ 298 = $-$ 3267.6 kJ mol $-$ 1

Q6

500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm. The final temperature and the change in internal energy respectively are: Given: R = 8.3 J K-1 mol-1

A 368 K and 500 J

B 348 K and 300 J

C 378 K and 300 J

D 378 K and 500 J

Correct Answer

Option B

Solution

Amount of energy transpired as hat, $Q=500 \mathrm{~J}$ Number of moles of Argon, $n=0.5 \mathrm{~mol}$ Temperature (Initial), $T_i=298 \mathrm{~K}$ Pressure, $P=1.00 \mathrm{~atm}$ Final temperature, $T_F=$ ?

Change in internal energy, $\Delta V=$ ?

R=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

For a constant pressure process, the heat added is related to the temperature changes as $Q=n C_p \Delta T$ ....... (1)

\begin{aligned} &\text{ Also, the change in internal energy formula is }\\ &\Delta U=n C_V \Delta T \quad-(2)\quad\text{.... (2)} \end{aligned}

\begin{aligned} &\text{ For monoatomic gas (Argon), }\\ &\begin{aligned} C_v & =\frac{3}{2} R \\ \text{ So, } C_p & =C_v+R \quad\left(C_p-C_v=R\right) \\ & =\frac{3}{2} R+R=\frac{5}{2} R \end{aligned} \end{aligned}

\begin{aligned} &\text{ In equation (1) } \Rightarrow\\ &\begin{aligned} & Q=n C_p \Delta T \\ & Q=n C_p\left(T_F-T_i\right) \\ & 500 \mathrm{~J}=0.5 \mathrm{~mol} \times \frac{5}{2} R\left(T_F-T_i\right) \quad R=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ & 500 \mathrm{~J}=0.5 \mathrm{~mol} \times \frac{5}{2} \times 8.3 \mathrm{~J} K^{-1} \mathrm{~mol} ^{-1}\left(T_F-T_i\right) \end{aligned} \end{aligned}

$T_F-T_i=\dfrac{500 \mathrm{~J}}{0.5 \mathrm{mol} \times \dfrac{5}{2} \times 8.3 \text{ J K}^1 \mathrm{~mol}^{-1}}$

\begin{aligned} & =\frac{500}{0.5 \times \frac{5}{2} \times 8.3} \mathrm{k} \\ & =48.2 \mathrm{k} \\ T_F & =48.2 \mathrm{k}+298 \mathrm{k} \\ & =346.2 \mathrm{k} \\ & \approx 348 \mathrm{k} \end{aligned}

\begin{aligned} &\text{ In equation (2) } \Rightarrow\\ &\begin{aligned} \Delta U & =n C_V \Delta T \\ & =0.5 \mathrm{~mol} \times \frac{3}{2} R \times 48.2 \mathrm{~K} \\ & =0.5 \mathrm{~mol} \times \frac{3}{2} \times 8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol^{-1}} \times 48.2 \mathrm{~K} \\ & =300.045 \mathrm{~J} \approx 300 \mathrm{~J} \end{aligned} \end{aligned}

The final temperature is, 348 k change in internal energy is 300 J So, correct answer is Option 2) 348 K and 300 J

Q7

During which of the following processes, does entropy decrease? (A) Freezing of water to ice at 0

^\circ

C (B) Freezing of water to ice at

-

10

^\circ

C (C) N2(g) + 3H2(g)

\to

2NH3(g) (D) Adsorption of CO(g) on lead surface. (E) Dissolution of NaCl in water Choose the correct answer from the options given below :

A (A), (C) and (E) only

B (B) and (C) only

C (A), (B), (C) and (D) only

D (A) and (E) only

Correct Answer

Option C

Solution

A, B $\to$ Freezing of water will decrease entropy as particles will move closer and forces of attraction will increase.

This leads to decrease in randomness.

So entropy decrease.

C $\to$ No. of molecules decreasing D $\to$ Adsorption will lead to decrease in randomness of gaseous particles.

E $\to$ NaCl(s) $\to$ Na+(aq) + Cl–(aq)

\Delta

S > 0 So, (A, B, C, D) decreases entropy.

Q8

At 25

^\circ

C and 1 atm pressure, the enthalpy of combustion of benzene (I) and acetylene (g) are

-

3268 kJ mol

-

1 and

-

1300 kJ mol

-

1, respectively. The change in enthalpy for the reaction 3 C2H2(g)

\to

C6H6 (I), is :

A +324 kJ mol

-

1

B +632 kJ mol

-

1

C

-

632 kJ mol

-

1

D

-

732 kJ mol

-

1

Correct Answer

Option C

Solution

I.

$\mathrm{C}_{6} \mathrm{H}_{6}(\ell)+\dfrac{15}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$

\Delta \mathrm{H}_{1}=-3268 \mathrm{~kJ} / \mathrm{mol}

II.

$\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+\dfrac{5}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$

\Delta \mathrm{H}_{2}=-1300 \mathrm{~kJ} / \mathrm{mol}

III.

$3 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{6}(\ell) \quad \Delta \mathrm{H}_{3}$ Applying Hess's law of constant heat summation

\begin{aligned} \Delta \mathrm{H}_{3} &=3 \times \Delta \mathrm{H}_{2}-\Delta \mathrm{H}_{1} \\\\ &=3 \times(-1300)-(-3268) \\\\ &=-632 \mathrm{~kJ} / \mathrm{mol} \end{aligned}

Q9

Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non-expansion work is zero)

A Adiabatic process :

\Delta

U= – w

B Cyclic process : q = –w

C Isochoric process :

\Delta

U= q

D Isothermal process : q = – w

Correct Answer

Option A

Solution

From 1st law of thermodynamics we know,

\Delta

U = q + W Option A : In adiabatic process exchage of heat = 0 $\therefore$ q = 0 $\therefore$ From 1st law of thermodynaics,

\Delta

U = W So option A is wrong.

Option B : U is a state function.

In cyclic process, initial state and final state both are same.

So change in all the state function in cyclic process will be zero.

$\therefore$

\Delta

U = 0 $\therefore$ From 1st law of thermodynaics, q + W = 0 $\Rightarrow$ q = -W So option B is correct..

Option C : In isochoric process volume (V) is constant.

So dV = 0.

We know, W =

- \int {{P_{ex}}} dV

$\therefore$ W = 0 $\therefore$ From 1st law of thermodynaics,

\Delta

U = q So option C is correct. Option D : In isothermal process temerature (T) is constant. So dT = 0. We know,

\Delta

U = nCvdT $\therefore$

\Delta

U = 0 $\therefore$ From 1st law of thermodynaics, q + W = 0 $\Rightarrow$ q = -W So option D is correct.

Q10

One mole of an ideal gas expands isothermally and reversibly from

10 \mathrm{dm}^3

to

20 \mathrm{dm}^3

at 300 K .

\Delta \mathrm{U}, \mathrm{q}

and work done in the process respectively are Given:

\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

\ln 10=2.3

\log 2=0.30

\log 3=0.48

A

0,21.84 \mathrm{~kJ},-1.726 \mathrm{~J}

B

0,21.84 \mathrm{~kJ}, 21.84 \mathrm{~kJ}

C

0,1.718 \mathrm{~kJ},-1.718 \mathrm{~kJ}

D

0,-17.18 \mathrm{~kJ}, 1.718 \mathrm{~J}

Correct Answer

Option C

Solution

\begin{aligned} & (10 \mathrm{~L}, 300 \mathrm{~K}) \xrightarrow{\mathrm{n}=1}(20 \mathrm{~L}, 300 \mathrm{~K}) \\ & -\mathrm{q}=\mathrm{w}=-\mathrm{nRT} \ln \frac{\mathrm{~V}_2}{\mathrm{~V}_1} \\ & =-8.3 \times 300 \times \ln \left(\frac{20}{10}\right) \\ & =-1.718 \mathrm{~kJ} \\ & \Rightarrow \mathrm{q}=1.718 \mathrm{~kJ} \\ & \mathrm{w}=-1.718 \mathrm{~kJ} \\ & \Delta \mathrm{U}=0(\because \Delta \mathrm{~T}=0) \end{aligned}