Thermodynamics — Page 6 | JEE Chemistry

Q51

A gas undergoes change from state A to state B. In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another process during which 3 J of heat is evolved. In this reverse process of B to A :

A 10 J of the work will be done by the gas.

B 6 J of the work will be done by the gas.

C 10 J of the work will be done by the surrounding on gas.

D 6 J of the work will be done by the surrounding on gas.

Correct Answer

Option D

Solution

From first law of thermodynamics, we have

\Delta

U = q + w $\bullet$ For state A to B : q = +5 J, w = $-$ 8 J and

\Delta

UAB = 5 + ( $-$ 8) = $-$ 3 J $\bullet$ For state B to A : q = $-$ 3 J Since the internal energy is a state function, we have

\Delta

UBA = $-$

\Delta

UAB. Therefore,

\Delta

UBA = q + w 3 = $-$ 3 + w $\Rightarrow$ w = 6 J Positive value of work indicates that 6 J of work is done by the surrounding on gas.

Q52

An ideal gas undergoes isothermal expansion at constant pressure. During the process :

A enthalpy increases but entropy decreases.

B enthalpy remains constant but entropy increases.

C enthalpy decreases but entropy increases.

D Both enthalpy and entropy remain constant.

Correct Answer

Option B

Solution

In an isothermal expansion process, the temperature of the system remains constant throughout the process.

Since, for an ideal gas, U depends only on temperature, we have

\Delta

U = 0. The enthalpy change of the system in isothermal expansion is also zero as

\Delta

H =

\Delta

U + nR

\Delta

T = 0 + 0 = 0 The entropy change foe an isothermal process is given by

\Delta S = nR\ln \left( {{{{V_f}} \over {{V_i}}}} \right) \ge 0

Q53

\Delta

U is equal to :

A Isobaric work

B Adiabatic work

C Isothermal work

D Isochoric work

Correct Answer

Option B

Solution

From 1st law of thermodynamics ΔU = q + w For adiabatic process : q = 0 $\therefore$ ΔU = w So change in internal energy (ΔU) is equal to adiabatic work.

Q54

For which of the following reactions,

\Delta

H is equal to

\Delta

U?

A N2(g) + 3H2(g)

\to

2NH3(g)

B 2HI(g)

\to

H2(g) + I2(g)

C 2NO2(g)

\to

N2O4(g)

D 2SO2(g) + O2(g)

\to

2SO3(g)

Correct Answer

Option B

Solution

\Delta

H =

\Delta

$\mu$ +

\Delta

ng RT

\therefore\,\,\,

\Delta

H =

\Delta

$\mu$ When,

\Delta

ng RT = 0 $\Rightarrow$

\Delta

ng = 0 For this reaction, 2HI(g) $\to$ H2 (g) + I2 (g)

\Delta

ng = (1 + 1) $-$ 2 = 0 $\therefore$

\Delta

H =

\Delta

$\mu$

Q55

Among the following, the set of parameters that represents path function, is : (A) q + w (B) q (C) w (D) H–TS

A (B) and (C)

B (A) and (D)

C (B), (C) and (D)

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

Correct Answer

Option A

Solution

(A) q + w =

\Delta

E, state function (B) q, Path function (C) w, Path function (D) H – TS = G, State function

Q56

During compression of a spring the work done is 10kJ and 2kJ escaped to the surroundings as heat. The change in internal energy,

\Delta

U(inkJ) is :

A - 12

B 8

C - 8

D 12

Correct Answer

Option B

Solution

Here heat is released so q is negative. $\therefore$ q = - 2 kJ Work done on the system, w = 10 kJ From first law of thermodynamics,

\Delta

U = q + w = -2 + 10 = 8 kJ

Q57

A process will be spontaneous at all temperatures if :

A

\Delta

H < 0 and

\Delta

S > 0

B

\Delta

H < 0 and

\Delta

S < 0

C

\Delta

H > 0 and

\Delta

S < 0

D

\Delta

H > 0 and

\Delta

S > 0

Correct Answer

Option A

Solution

A reaction is spontaneous if

\Delta

G is negative. and we know that

\Delta

G =

\Delta

H – T

\Delta

S If

\Delta

H = –ve and

\Delta

S = + ve then at all the temperature the process will be spontaneous.

Q58

Enthalpy of sublimation of iodine is 24 cal g–1 at 200 oC. If specific heat of I2(s) and l2 (vap) are 0.055 and 0.031 cal g–1K –1 respectively, then enthalpy of sublimation of iodine at 250 oC in cal g–1 is :

A 2.85

B 22.8

C 11.4

D 5.7

Correct Answer

Option B

Solution

Kirchoff's Equation :

{{\Delta {H_{{T_2}}} - \Delta {H_{{T_1}}}} \over {{T_2} - {T_1}}} = \Delta \left( {{C_p}} \right)

Sublimation of iodine : I2(s) $\to$ I2(g)

{{\Delta {H_{250^\circ C}} - \Delta {H_{200^\circ C}}} \over {250 - 200}} = {C_p}\left[ {{I_2}(g)} \right] - {C_p}\left[ {{I_2}(s)} \right]

{{\Delta {H_{250^\circ C}} - 24} \over {50}} = \left( {0.031 - 0.055} \right)

$\Rightarrow$

{\Delta {H_{250^\circ C}}}

= 22.8 cal g–1

Q59

An ideal gas is allowed to expand form 1 L to 10 L against a constant external pressure of I bar. The work done in kJ is :

A +10.0

B –0.9

C – 2.0

D – 9.0

Correct Answer

Option B

Solution

This is an irreverseable process as gas is expanding against a constant external process.

Work done in irreverseable process W = - Pext

\Delta

V = - 1 bar $\times$ 9 L = - 105 Pa $\times$ 9 $\times$ 10-3 m3 = - 9 $\times$ 102 N-m = - 900 J = - 0.9 kJ

Q60

An ideal gas undergoes isothermal compression from 5m3 to 1 m3 against a constant external pressure of 4 Nm–2. Heat released in this process is used to increase the temperature of 1 mole of Al. If molar heat capacity of Al is 24 J mol–1 K–1, the temperature of Al increases by :

A

{2 \over 3}K

B

{3 \over 2}K

C 1 K

D 2 K

Correct Answer

Option A

Solution

Work done on isothermal irreversible for ideal gas = $-$ Pext (V2 $-$ V1) = $-$ 4 N/m2 (1 m3 $-$ 5m3) = 16 Nm Isothermal process for ideal gas

\Delta

U = 0 q = $-$ w = $-$ 16 Nm = $-$ 16 J Heat used to increase temperature of Al q = n Cm

\Delta

T 16 J = 1 $\times$ 24

{J \over {mol.K}}

$\times$

\Delta

T

\Delta

T =

{2 \over 3}

K