Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Thermodynamics

Multiple Choice Questions

101.

Two moles of an ideal monoatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

(a) 195 K (b) 2.7 kJ
(a) 189 K (b) 2.7 kJ
(a) 195 K (b) –2.7 kJ
(a) 189 K (b) – 2.7 kJ

(a) 189 K (b) – 2.7 kJ

In an adiabatic Process

$T V^{γ - 1} = C o n s \tan t O r T_{1} {V_{1}}^{γ - 1} = T_{2} {V_{2}}^{γ - 1} F o r m o n o a t o m i c g a s γ = \frac{5}{3} (300) V^{2 / 3} = T_{2} (2 V)^{2 / 3} \Rightarrow T_{2} = \frac{300}{(2)^{2 / 3}} T_{2} = 189 K (f i n a l t e m p e r a t u r e) C h a n g e i n i n t e r n a l e n e r g y, ∆ U = n \frac{f}{2} R ∆ T = 2 (\frac{3}{2}) (\frac{25}{3}) (- 111) = - 2.7 k J$

102.

For the process A (1, 0.05 atm, 32°C) → A (g, 0.05 atm, 32°C)

The correct set of thermodynamic parameters is

$∆$ G = 0 and $∆$ S = -ve
$∆$ G = 0 and $∆$ S = +ve
$∆$ G = +ve and $∆$ S = 0
$∆$ G = -ve and $∆$ S = 0

$∆$ G = 0 and $∆$ S = +ve

(i) Since, n and T are constants.

$∴ ∆$ G = 0

Hence, the system is equilibrium.

(ii) Since, P → decreases

$∴ ∆$ S → increases, i.e. it is +ve

Therefore, among all the options given, the correct answer is in option b, i.e. $∆$ G = 0 and $∆$ S = +ve.

103.

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 250C; the heat of combustion (in kJ mol^–1) of benzene at constant pressure will be (R = 8.314 JK^–1 mol^–1)

–3267.6
4152.6
–452.46
3260

–3267.6

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

Δ_ng = 6 - 7.5
= -1.5 (change in gaseous mole)
ΔU or ΔE = - 3263.9 kJ
ΔH = ΔU + Δn_gRT
Δ_ng = - 1.5
R = 8.314 JK^-1 mol^-1
T = 298 K
So ΔH = -3263.9 + (-1.5) 8.314 x 10^-3 x 298
= -3267.6 kJ
ΔH = Heat at constant pressure
ΔU/ΔE = Heat at constant volume
R = gas constant

104.
The first law of thermodynamics for isothermal process is

q = -W

$∆$ U = W

$∆$ U = q_v

$∆$ U = -q_v

q = -W

According to first law of thermodynamics,

$∆$ U = q + W

where, $∆$ U = internal energy; q = Heat and W = work done

For isothermal process,

$∆$ T = 0; $∆$ U = 0

Therefore, q = -W

105.

Identify the invalid equation

$∆ H = {ΣH}_{products} - {ΣH}_{reactants}$
$∆ H = ∆ U + p ∆ V$
$∆ H °_{(reaction)} = ΣH °_{(products bonds)} - ΣH °_{(reactant bonds)}$
$∆ H = ∆ U + ∆ nRT$

$∆ H °_{(reaction)} = ΣH °_{(products bonds)} - ΣH °_{(reactant bonds)}$

Relation between heat of reaction ( $∆$ rH° ) and bond enthalpies of reactants and products is

$∆ rH ° = {ΣBE}_{reactants} - {ΣBE}_{products} ∴ ∆ H °_{reaction} = ΣH °_{product bonds} - ΣH °_{reactant bonds}$

106.

Which of the following lines correctly show the temperature dependence of equilibrium constant K, for an exothermic reaction?

A and D
A and B
B and C
C and D

A and B

$∆ G^{°} = - R T \ln K ∆ H ° - T ∆ S ° = - R T \ln K - \frac{∆ H °}{R T} + \frac{∆ S °}{R} = \ln K$

Therefore ln K vs 1/T the graph will be a straight line with slope equal to $- \frac{∆ H^{°}}{R}$ .Since reaction is
exothermic, therefore $∆ H^{°}$ itself will be negative resulting in positive slope.

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

104. The first law of thermodynamics for isothermal process is q = -W ∆U = W ∆U = qv ∆U = -qv

104.
The first law of thermodynamics for isothermal process is

q = -W

$∆$ U = W

$∆$ U = q_v

$∆$ U = -q_v