Question

One mole of an ideal monoatomic gas is taken round cyclic process $ABCA$ as shown in figure. Calculate the maximum temperature attained by the gas during the cycle?

• A. $3T_1$
• B. $2T_1$
• C. $4T_1$
• D. $5T_1$

Answer 1

The correct option is C $4T_1$

Maximum temperature will lie between $B$ and $C$.

$PV$ equation for the product $BC$ can be written as

$P=-mV+C$

here $m=\frac{2P_o}{2V_o}=\frac{P_o}{V_o}andC\Rightarrow \frac{y-3P_o}{V_o-0}\frac{P_o}{V_o}=\Rightarrow y=4P_{o}C=4P_o$

$P=-\left(\frac{P_o}{V_o}\right)V+{4P}_o\left(1\right)PV=-\left(\frac{P_o}{V_o}\right)V^2+{4P}_{o}VRT=-\left(\frac{P_o}{V_o}\right)V^2+{4P}_{o}V[n=1\Rightarrow PV=RT]T=\frac{1}{R}[{4P}_{o}V-\left(\frac{P_o}{V_o}\right)V^2]\left(2\right)$

For $T$ to be maximum, $\frac{dT}{dV}=0$

${4P}_o-2\frac{P_o}{V_o}V=0{4P}_o=2\frac{P_o}{V_o}VV=2V_o$

Substituting it in equation (2), we get

$T_{max}=\frac{1}{R}[{4P}_o\left(2V_o\right)-\left(\frac{P_o}{V_o}\right){\left(2V_o\right)}^2]=\frac{1}{R}[{8P}_o{V}_o-{4P}_o{V}_o]T_{max}=\frac{{4P}_o{V}_o}{R}=4T_1$