One mole of an ideal gas undergoes a reversible process T = T0 + αV, where T0 and α are constants. I

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One mole of an ideal gas undergoes a reversible process T = T0 + αV, where T0 and α are constants. If its volume increases from V1 to V2, then the amount of heat transferred to the gas is

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$C_{m} = C_{V,m} + \frac{P.dV}{dT} = C_{V,m} + \frac{P}{\alpha} = C_{V,m} + \frac{RT}{V.\alpha}$

$= C_{V,m} + \frac{R\left( T_{0} + \alpha V \right)}{V\alpha} = C_{P,m} + \frac{RT_{0}}{V\alpha}$

Now, $q = \int_{T_{1}}^{T_{2}}{C_{m}.dT} = \int_{V_{1}}^{V_{2}}{C_{m}.(\alpha.dV)}$

$= \int_{V_{1}}^{V_{2}}{\left( C_{P,m}.\alpha + \frac{RT_{0}}{V} \right)dV} $$$= \alpha.C_{P,m}\left( V_{2} - V_{1} \right) + RT_{0}.\ln\frac{V_{2}}{V_{1}}$$

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