Question

The molar heat capacity C for an ideal gas going through a process is given by $C=\frac{a}{T}$, where 'a' is a constant. If $\gamma =\frac{C_P}{C_V}$, the work done by one mole of gas during heating from $T_0$ and $\eta T_0$ will be:

• A. $aln\left(\eta \right)$
• B. $\frac{1}{aln\left(\eta \right)}$
• C. $aln\left(\eta \right)-\left(\frac{\eta -1}{\gamma -1}\right)R{T}_0$
• D. $aln\left(\eta \right)-(\gamma -1)R{T}_0$

Answer 1

The correct option is C $aln\left(\eta \right)-\left(\frac{\eta -1}{\gamma -1}\right)R{T}_0$

Q=w+Au

w=Q-Au Q=cdt (for 1 mole)
Q=${\int }_{T0}^{T1}\left[\frac{a}{t}dT\right]=a\ln n$

$Au=\eta \frac{R}{\gamma -1}At$ (here n=1mole)

$Au=\frac{R}{\gamma -1}{T}_0(\gamma -1)$

$w=a\ln \left(\gamma \right)-\left(\frac{\eta -1}{\gamma -1}\right)R{T}_0$