Question

A monoatomic gas at a pressure P, having a volume V expands isothermally to a volume 2V and then adiabatically to a volume 16V. The final pressure of the gas is: (Take $\gamma =\frac{5}{3}$)

Answer 1

Step 1: Given

Pressure is P.

Volume is V.

After isothermal expansion, the volume is 2V.

After adiabatic expansion, volume is 16 V.

Pressure after the isothermal expansion is P' (say).

Pressure after the adiabatic expansion is P_f (say).

Step 2: Formula Used

For isothermal expansion: $PV=cons\tan t$

For adiabatic expansion: $P{V}^{\gamma }=cons\tan t$

Step 3: Solution:

Find P' using the formula $PV=cons\tan t$

$$\begin{gathered} PV=P\text{'}\left(2V\right) \\ \Rightarrow P=2P\text{'} \\ \Rightarrow P\text{'}=\frac{P}{2} \end{gathered}$$

Find P' using the formula $P{V}^{\gamma }=cons\tan t$

$$\begin{gathered} P\text{'}{\left(2V\right)}^{\gamma }=P_f{\left(16V\right)}^{\gamma } \\ \Rightarrow \frac{P}{2}{\left(2V\right)}^{\frac{5}{3}}=P_f{\left(16V\right)}^{\frac{5}{3}} \\ \Rightarrow P_f=\frac{\frac{P}{2}{\left(V\right)}^{\frac{5}{3}}}{{\left(8V\right)}^{\frac{5}{3}}} \\ \Rightarrow P_f=\frac{P}{2}{\left(\frac{1}{2^3}\right)}^{\frac{5}{3}} \\ \Rightarrow P_f=\frac{P}{2^6}=\frac{P}{64} \end{gathered}$$

Hence, the final pressure is $\frac{P}{64}$.