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Thermodynamics

Multiple Choice Questions

101.
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.

102.

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

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

103.

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

104.

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.

105.

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

106.

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

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

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

101. 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

101.
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