Question

The density ($\rho$) Vs pressure (P) graph of an ideal gas (monoatomic) undergoing a cyclic process is shown in the figure The gas taken has molecular weight M and one mole of gas is taken The heat rejected by gas in process $3$ to $1$ in the cycle is

• A. $\frac{3P_{0}M}{{\rho }_0}$
• B. $\frac{3P_{0}M}{2{\rho }_0}$
• C. $\frac{5P_{0}M}{2{\rho }_0}$
• D. $\frac{5P_{0}M}{3{\rho }_0}$

Answer 1

The correct option is B $\frac{3P_{0}M}{2{\rho }_0}$

Ideal gas equation at 1 : $P_{o}M={\rho }_{o}R{T}_1⟹T_1=\frac{P_{o}M}{{\rho }_{o}R}$

Process 3 $\to$ 1 is an isochoric process : $\frac{P_1}{T_1}=\frac{P_3}{T_3}$

$\therefore$ $\frac{P_o}{\left(\frac{P_{o}M}{{\rho }_{o}R}\right)}=\frac{2P_o}{T_3}⟹T_3=\frac{2P_{o}M}{{\rho }_{o}R}$

For monoatomic gas, $C_V=\frac{3}{2}R$

$\therefore$ Heat rejected at constant volume $Q_v=\Delta U=n{C}_v(T_1-T_3)$

$\therefore$ $Q_v=1\times \frac{3}{2}R(\frac{P_{o}M}{{\rho }_{o}R}-\frac{2P_{o}M}{{\rho }_{o}R})=-\frac{3}{2}\frac{P_{o}M}{{\rho }_o}$ (-ve sign implies that heat is rejected)