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Q

A monoatomic gas at a pressure P, having a volume V expands isothermally to a volume 2 V, and then adiabatically to a volume 16 V. the final pressure of the gas is:

$take\:\:\left( \gamma= \frac{5}{3} \right )$

• Option 1)

64 P

• Option 2)

32 P

• Option 3)

$\frac{P}{64}$

• Option 4)

16 P

83 Views

As learnt in

Equation of state -

$dQ= 0$

$n\, C_{V}\, dT+PdV= 0$

- wherein

On solving

$\gamma \frac{dV}{V}+\frac{dP}{P}= 0$

$\Rightarrow PV^{\gamma }= constant$

For Isothermal expansion,

$P_{i}= P, \ V_{i}= V,\ V_{1}= 2V\ \ \Rightarrow P_{1} =P_{i}\cdot \frac{V}{2V}= \frac{P}{2}$

$P_{1}= \frac{P}{2},\ V_{1}=2V,\ V_{f}= 16V,\ P_{f} =?$

$PV^{\gamma }= \ constant$

$\Rightarrow P_{1}V_{1}^{\gamma } = P_{f}V_{f}^{\gamma }\ \ \Rightarrow P_{f}= P_{1}\cdot \left ( \frac{V_{1}}{V_{f}} \right )^{\gamma }$

$P_{f}= \frac{P}{2}\cdot \left ( \frac{2V}{16V} \right )^{\frac{5}{3}} = \frac{P}{2}\cdot \left ( \frac{1}{2} \right )^{5} = \frac{P}{64}$

Option 1)

64 P

This option is incorrect

Option 2)

32 P

This option is incorrect

Option 3)

$\frac{P}{64}$

This option is correct

Option 4)

16 P

This option is incorrect

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