Question

The equation of state for a gas is given by $PV = nRT + \alpha V$, where n is the number of moles and a is a positive constant. The initial temperature and pressure of one mole of the gas contained in a cylinder are $T_o$ and $P_o$ respectively. The work done by the gas when its temperature doubles isobarically will be :

• A. $\frac{P_{O} T_{O}R}{P_{O} -\alpha}$
• B. $\frac{P_{O} T_{O}R}{P_{O} +\alpha}$
• C. $P_{O} T_{O}R \,\ell n\, 2$
• D. $P_{O} T_{O}R $

Answer 1

Answer: (A)

$P_{0}V_{0}=nRT_{0}$
$P_{0}V_{0}=nRT$
$T_{f} =2T_{0}$
$W=\int PdV
=\int \left(\frac{nRT}{V}+\alpha\right)dv$
$PV=nRT+\alpha V$
$\int pdV=$$\int\limits^{2T_0}_{{T_0}}nRdT+\int\limits^{V_f}_{{V_i}}\alpha dV$
$=nRT_0+ \alpha \, V_i$
$=nRT_{0}+\alpha\left(\frac{nRT_{0}}{P_{0}}\right)$
$=nRT_{0}\left(1+\frac{\alpha}{P_{0}}\right)
PV = nRT + \alpha V$
$\int pdV=\int nRdT+\int\alpha dV$
$W=nRT_{0}+\alpha\left[\frac{nRT_{0}}{P_{0}-\alpha}\right]$
$W=nRT_{0}\left[1+\frac{\alpha}{p_{0}-\alpha}\right]$
$=nR_{0}T_{0}\left[\frac{P_{0}}{P_{0}-\alpha}\right]$
$=\frac{nRT_{0}\,P_{0}}{P_{0}-\alpha}$