Question

Two moles of an ideal monoatomic gas, initially at pressure $P_1$ and volume $V_1$ undergo an adiabatic compression until its volume is $V_2$. Then the gas is given heat Q at constant volume $V_2$. Sketch the complete process on a p-V diagram.

Answer 1

For an adiabatic process, $P{v}^y=$ Constant

and $y=\frac{5}{3}$ for a monoatomic gas

$P_1{v}_1^{5/3}=P_2{v}_2^{5/3}$

As, it is a compression process, $v_2<v_1$ and $v_2<v_1$ and so, $P_2>P_1$

Now, the gas is given heat ${}^{\prime }{Q}^{\prime }$ at constant $v_2.$

By first law of thermodynamics, $dq=dv+dw$

$\Rightarrow Q=\frac{t}{2}nR△T+0$

and $Pv=nRT$ $\Rightarrow Q=\frac{t}{2}nR\frac{v△P}{nR}-\frac{t}{2}v△P$

$\Rightarrow △PV=△nRT$ $\Rightarrow \frac{2Q}{f{v}_2}=p_3-p_2$

$△Pv=△nRT$ $\Rightarrow p_3=p_2+\frac{2Q}{3v_2}$

$v△P=nR△T$

$\Rightarrow △T=\frac{v△P}{nR}$

Hence, solved.