Practice Start Test

Thermodynamics

NEET Chemistry · 93 questions · Page 2 of 10 · Click an option or "Show Solution" to reveal answer

Q11
A vessel contains 3.2 g of dioxygen gas at STP (273.15 K and 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes one third of the original pressure. The volume of new vessel in L is : (Given : molar volume at STP is 22.4 L)
A 67.2
B 6.72
C 2.24
D 22.4
Correct Answer
Option B
Solution

At constant temperature and amount

P1V1=P2V2{P_1}{V_1} = {P_2}{V_2}
P1V1=P13V2{P_1}{V_1} = {{{P_1}} \over 3}{V_2}

[∴\therefore

P2=P13{P_2} = {{{P_1}} \over 3}

]

V2=3V1{V_2} = 3{V_1}

mole of

O2(g)=3.232=0.1{O_2}(g) = {{3.2} \over {32}} = 0.1

mole Volume of

O2(g)=(0.1×22.4){O_2}(g) = (0.1 \times 22.4)

L = 2.24 L At STP (V 1 )

V2=3V1=3×2.24{V_2} = 3{V_1} = 3 \times 2.24

= 6.72 L

Q12
Which one among the following is the correct option for right relationship between C P and C V for one mole of ideal gas?
A C V = RC P
B C P + C V = R
C C P −- C V = R
D C P = RC V
Correct Answer
Option C
Solution

At constant volume, q V = C V

Δ\Delta

T =

Δ\Delta

U At constant pressure, q P = C P

Δ\Delta

T =

Δ\Delta

H For a mole of an ideal gas,

Δ\Delta

H =

Δ\Delta

U +

Δ\Delta

(PV) =

Δ\Delta

U +

Δ\Delta

(RT) =

Δ\Delta

U + R

Δ\Delta

T On putting the values of

Δ\Delta

H and

Δ\Delta

U, we have C P

Δ\Delta

T = C V

Δ\Delta

T + R

Δ\Delta

T C P = C V + R C P −- C V = R

Q13
For irreversible expansion of an ideal gas under isothermal condition, the correct option is :
A Δ\DeltaU ≠\ne 0, Δ\DeltaS total = 0
B Δ\DeltaU = 0, Δ\DeltaS total = 0
C Δ\DeltaU ≠\ne 0, Δ\DeltaS total ≠\ne 0
D Δ\DeltaU = 0, Δ\DeltaS total ≠\ne 0
Correct Answer
Option D
Solution

∙\bullet For a spontaneous process,

Δ\Delta

S total > 0 and since irreversible process is always spontaneous therefore

Δ\Delta

S total > 0. ∙\bullet Since

Δ\Delta

U = nC V

Δ\Delta

T and

Δ\Delta

T = 0 for isothermal process therefore

Δ\Delta

U = 0.

Q14
Hydrolysis of sucrose is given by the following reaction. Sucrose + H 2 O ⇌ Glucose + Fructose If the equilibrium constant (K c ) is 2 × \times 10 13 at 800 K, the value of ΔrGΘ\Delta r{G^\Theta } at the same temperature will be :
A 8.314 J mol−1K−1×300K×In(2×1013)mo{l^{ - 1}}{K^{ - 1}} \times 300K \times In(2 \times {10^{13}})
B 8.314 J mol−1K−1×300K×In(3×1013)mo{l^{ - 1}}{K^{ - 1}} \times 300K \times In(3 \times {10^{13}})
C − - 8.314 J mol−1K−1×300K×In(4×1013)mo{l^{ - 1}}{K^{ - 1}} \times 300K \times In(4 \times {10^{13}})
D -8.314 J mol−1K−1×300K×In(2×1013)mo{l^{ - 1}}{K^{ - 1}} \times 300K \times In(2 \times {10^{13}})
Correct Answer
Option D
Solution
ΔG\Delta G

=

ΔG\Delta G
∘^\circ

+ RT ln Q At equilibrium

ΔG\Delta G

= 0, Q = K eq So,

ΔrG∘{\Delta _r}G^\circ

= −- RT in K eq

ΔrG∘{\Delta _r}G^\circ

= −- 8.314 J mol -1 K -1 ×\times 300K ×\times ln (2 ×\times 10 13 )

Q15
For the reaction, 2Cl(g) → \to Cl 2 (g), the correct option is :
A ΔrH>0{\Delta _r}H > 0 and ΔrS<0{\Delta _r}S < 0
B ΔrH<0{\Delta _r}H < 0 and ΔrS>0{\Delta _r}S > 0
C ΔrH<0{\Delta _r}H < 0 and ΔrS<0{\Delta _r}S < 0
D ΔrH>0{\Delta _r}H > 0 and ΔrS>0{\Delta _r}S > 0
Correct Answer
Option C
Solution

Given reaction, 2Cl(g)→\toCl 2 (g) This is an endothermic reaction because it requires energy to break bonds.

So, the reverse reaction is exothermic

ΔrH<0{\Delta _r}H < 0

Also, two gaseous atoms combine together to form 1 gaseous molecule. So, the randomness,

ΔrS<0{\Delta _r}S < 0
Q16
The correct option for free expansion of an ideal gas under adiabatic condition is :
A q = 0, Δ\Delta T < 0 and w > 0
B q < 0, Δ\Delta T = 0 and w = 0
C q > 0, Δ\Delta T > 0 and w > 0
D q = 0, Δ\Delta T = 0 and w = 0
Correct Answer
Option D
Solution

Free expansion, so p ex = 0 So, w = −- p ex

Δ\Delta

V = 0 Since, adiabatic process, so q = 0 Since both q and w are equal to zero.

Then according to first law of thermodynamics

Δ\Delta

E = q + w

Δ\Delta

U = 0 Hence,

Δ\Delta

T = 0

Q17
In which case change in entropy is negative?
A Sublimation of solid to gas
B 2H(g) → \to H 2 (g)
C Evaporation of water
D Expansion of a gas at temperature
Correct Answer
Option B
Solution

2H(g) →\to H 2 (g) Due to bond formation, entropy decreases.

Q18
Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is [Given that 1 L bar = 100 J]
A 25 J
B 30 J
C -30 J
D 5 kJ
Correct Answer
Option C
Solution

W = –P ext (V 2 –V 1 ) P ext = 2 bar V 1 = 0.1 L V 2 = 0.25 L W = –2 bar[0.25 – 0.1] L W = –2 × 0.15 bar L W = –0.30 bar L W = (–0.30) × 100 = –30 J

Q19
The bond dissociation energies of X 2 , Y 2 and XY are in the ratio of 1 : 0.5 : 1. Δ\Delta H for the formation of XY is –200 kJ mol –1 . The bond dissociation energy of X 2 will be
A 800 kJ mol –1
B 200 kJ mol –1
C 400 kJ mol –1
D 100 kJ mol –1
Correct Answer
Option A
Solution

Let bond dissociation energies of X 2 , Y 2 and XY are x kJ mol –1 , 0.5x kJ mol –1 and x kJ mol –1 respectively.

12{1 \over 2}

X 2 +

12{1 \over 2}

Y 2 →\to XY;

Δ\Delta

H = -200 kJ mol -1

Δ\Delta

H = Σ(B.E) Reactant - Σ(B.E) Product ⇒\Rightarrow -200 =

[12×x+12×0.5x]\left[ {{1 \over 2} \times x + {1 \over 2} \times 0.5x} \right]

-

[1×x]\left[ {1 \times x} \right]

On solving, x = 800 kJ mol –1

Q20
A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy Δ\Delta U of the gas in joules will be
A −- 500 J
B −- 505 J
C + 505 J
D 1136.25 J
Correct Answer
Option B
Solution

w = - P ext

Δ\Delta

V = -2.5(4.50 - 2.50) ⇒\Rightarrow - 5 L atm = - 5 ×\times 1.01.325 J = - 506.625 J

Δ\Delta

U = q + w As, the container is insulted, thus q = 0 Hence,

Δ\Delta

U = w = -506.625 J

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