An ideal gas has a molar specific heat capacity $C_{v}$ at constant volume. The molar specific heat capacity of this gas as a function of its volume V

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is $(C_{v} + \frac{R}{αV})$ if the gas undergoes the process ${T=T}_{0} e^{αV}$

is $(C_{v} + \frac{2R}{αV})$ if the gas undergoes the process ${T=T}_{0} e^{αV}$

is ${C_{v} + \frac{R}{(1+αV)}}$ if the gas undergoes the process ${P=P}_{0} e^{αV}$

is $(C_{v} + \frac{R}{αV})$ if the gas undergoes process ${P=P}_{0} e^{αV}$

answer is A, C.

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Detailed Solution

According to first law of thermodynamics, $d Q = d U + P d V$ (for one mole)
Now molar specific heat is given by
$C = \frac{d Q}{d T} = \frac{d U + P d V}{d T} = \frac{C_{v} d T + (R T / V) d V}{d T} = C_{v} + [\frac{R T}{V}] [\frac{d V}{d T}] ........ (i)$
Given $T = T_{0} e^{α v} o r, d T = α T_{0} e^{α v} d V ∴ \frac{d V}{d T} = \frac{1}{α T_{0} e^{α v}}$
Substituting the value of (dV/dT) from eq. (ii) in eq. (i)
$C = C_{v} + [\frac{R T}{V}] [\frac{1}{α T_{0} e^{α v}}] C = C_{v} + \frac{R T_{0} e^{α v}}{α V T_{0} e^{α v}} = C_{v} + \frac{R}{α V} (c) G i v e n t h a t P = P_{0} e^{α v} \frac{R T}{V} = P_{0} e^{α v} o r T = \frac{P_{0}}{R} V e^{α v} N o w C = C_{v} + [\frac{R T}{V}] [\frac{d V}{d T}] = C_{v} + (P_{0} e^{α v}) [\frac{R}{P_{0} e^{α v} (1 + α V)}] = C_{v} = \frac{R}{(1 + α V)}$