An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation P = kV. Show that the molar heat capacity of the gas for the process is given by $C = C_{V} + \frac{R}{2}$ .

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Solution

For the gas, work done will be given by

$W = \int - P d V = - \int k V d V = - k (\frac{V_{2}^{2} - V_{1}^{2}}{2})$

Also, since $P = k V$ ,

$W = - \frac{P_{2} V_{2} - P_{1} V_{1}}{2} = - \frac{n R Δ T}{2}$

Also, internal energy change will be

$Δ U = n C_{V} Δ T$

Since $Δ U = Q + W$

$Q = n C Δ T = Δ U - W = n C_{V} Δ T + \frac{R}{2} n Δ T$

$C = C_{V} + \frac{R}{2}$

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An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation P = kV. Show that the molar heat capacity of the gas for the process is given by C=CV+R2.

For the gas, work done will be given by W=∫−PdV=−∫kVdV=−k(V22−V212)Also, since P=kV,W=−P2V2−P1V12=−nRΔT2Also, internal energy change will be ΔU=nCVΔTSince ΔU=Q+WQ=nCΔT=ΔU−W=nCVΔT+R2nΔTC=CV+R2

An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation P = kV. Show that the molar heat capacity of the gas for the process is given by $C = C_{V} + \frac{R}{2}$ .