One mole of an ideal gas with heat capacity at constant pressure Cp undergoes the process T-T0- V where T0 and - are constants- If its volume increases from V1 to V2- the amount of heat transfered to the gas is

From ideal gas equation, p=RTV=R(T_0+α V)V Work done by the gas W=∫_V_1^V_2p× dV=∫ _V_1^V_2R(T_0+α V)V× dV =RT_0lnV_2V_1+Rα (V_2 V_1) Change in internal energ ...

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Standard XII

Physics

Degree of Freedom

One mole of a...

Question

One mole of an ideal gas with heat capacity at constant pressure $C_{p}$ undergoes the process $T = T_{0} + α V$ where $T_{0}$ and $α$ are constants. If its volume increases from $V_{1}$ to $V_{2}$ , the amount of heat transfered to the gas is

C_{p} R T_{0} ln \frac{V_{2}}{V_{1}}

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α C_{p} (V_{2} - V_{1}) - R T_{0} ln \frac{V_{2}}{V_{1}}

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α C_{p} (V_{2} - V_{1}) + R T_{0} ln \frac{V_{2}}{V_{1}}

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R T_{0} \frac{V_{2}}{V_{1}} - α C_{p} (V_{2} - V_{1})

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Solution

The correct option is C $α C_{p} (V_{2} - V_{1}) + R T_{0} ln \frac{V_{2}}{V_{1}}$
From ideal gas equation,
$p = \frac{R T}{V} = \frac{R (T_{0} + α V)}{V}$

Work done by the gas
$W = \int_{V_{1}}^{V_{2}} p \times d V = \int_{V_{1}}^{V_{2}} \frac{R (T_{0} + α V)}{V} \times d V$
$= R T_{0} ln \frac{V_{2}}{V_{1}} + R α (V_{2} - V_{1})$

Change in internal energy
$Δ U = n C_{V} Δ T$
$= (C_{p} - R) [(T_{0} + α V_{2}) - (T_{0} + α V_{1})]$
$= α (C_{p} - R) (V_{2} - V_{1})$

Hence, amount of heat transferred $Q = W + Δ U$
$= R T_{0} ln \frac{V_{2}}{V_{1}} + α C_{p} (V_{2} - V_{1})$

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One mole of an ideal gas with heat capacity at constant pressure Cp undergoes the process T=T0+αV where T0 and α are constants. If its volume increases from V1 to V2, the amount of heat transfered to the gas is

One mole of an ideal gas with heat capacity at constant pressure $C_{p}$ undergoes the process $T = T_{0} + α V$ where $T_{0}$ and $α$ are constants. If its volume increases from $V_{1}$ to $V_{2}$ , the amount of heat transfered to the gas is