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# 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 $\gamma =\frac{5}{3}$)

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Solution

## Step 1: GivenPressure 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 Pf (say).Step 2: Formula UsedFor isothermal expansion: $PV=cons\mathrm{tan}t$For adiabatic expansion: $P{V}^{\gamma }=cons\mathrm{tan}t$Step 3: Solution:Find P' using the formula $PV=cons\mathrm{tan}t$$PV=P\text{'}\left(2V\right)\phantom{\rule{0ex}{0ex}}⇒P=2P\text{'}\phantom{\rule{0ex}{0ex}}⇒P\text{'}=\frac{P}{2}$Find P' using the formula $P{V}^{\gamma }=cons\mathrm{tan}t$$P\text{'}{\left(2V\right)}^{\gamma }={P}_{f}{\left(16V\right)}^{\gamma }\phantom{\rule{0ex}{0ex}}⇒\frac{P}{2}{\left(2V\right)}^{\frac{5}{3}}={P}_{f}{\left(16V\right)}^{\frac{5}{3}}\phantom{\rule{0ex}{0ex}}⇒{P}_{f}=\frac{\frac{P}{2}{\left(V\right)}^{\frac{5}{3}}}{{\left(8V\right)}^{\frac{5}{3}}}\phantom{\rule{0ex}{0ex}}⇒{P}_{f}=\frac{P}{2}{\left(\frac{1}{{2}^{3}}\right)}^{\frac{5}{3}}\phantom{\rule{0ex}{0ex}}⇒{P}_{f}=\frac{P}{{2}^{6}}=\frac{P}{64}$Hence, the final pressure is $\frac{P}{64}$.

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