The initial pressure and volume of a gas is ' P ' and ' V ' respectively. First by isothermal process gas is expanded to volume ' 9 ~V ' and then by adiabatic process its volume is compressed to ' V ' then its final pressure is (Ratio of specific heat at constant pressure to constant volume =3/2 )

Correct answer: (C) 3P. C_p/C_v=3/2 Case I: Isothermal process P_1 ~V_1=P_2 ~V_2 PV=P_2 × 9 ~V ∴ P_2=P/9 Case II: Adiabatic process P_2 ~V_2^γ=P_3 ~V_3^γ P/9(9 ~V)^γ=P_3(~V)^γ P_3= (P/9 ) 9^γ= (P/9 ) 9^3 / 2=P/9 × 27=3 P

Q101 / 150 · MHT CET 2021 (24 Sep Shift 1) · Physics

The initial pressure and volume of a gas is ' $\mathrm{P}$ ' and ' $\mathrm{V}$ ' respectively. First by isothermal process gas is expanded to volume ' $9 \mathrm{~V}$ ' and then by adiabatic process its volume is compressed to ' $\mathrm{V}$ ' then its final pressure is (Ratio of specific heat at constant pressure to constant volume $=\frac{3}{2}$ )

A
$6 \mathrm{~A}$
B
27P
Correct answer
C
3P
D
9P

Solution

$$ \frac{C_p}{C_v}=\frac{3}{2} $$ Case I: Isothermal process $$ \begin{aligned} & \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2 \\ & \mathrm{PV}=\mathrm{P}_2 \times 9 \mathrm{~V} \\ & \therefore \mathrm{P}_2=\frac{\mathrm{P}}{9} \end{aligned} $$ Case II: Adiabatic process $$ \begin{aligned} & \mathrm{P}_2 \mathrm{~V}_2^\gamma=\mathrm{P}_3 \mathrm{~V}_3^\gamma \\ & \frac{\mathrm{P}}{9}(9 \mathrm{~V})^\gamma=\mathrm{P}_3(\mathrm{~V})^\gamma \\ & \mathrm{P}_3=\left(\frac{\mathrm{P}}{9}\right) 9^\gamma=\left(\frac{\mathrm{P}}{9}\right) 9^{3 / 2}=\frac{\mathrm{P}}{9} \times 27=3 \mathrm{P} \end{aligned} $$

Hence, the correct answer is option: 'C'.

Solution

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