Question

A monoatomic gas at a pressure P having a volume v expands isothermally to a volume 3V and then compresses adiabatically to a volume v the final pressure of the gas is
a ) 3^ 5 / 3 P
b ) 3^ 2 / 3 P
c) P / 3
d) 3 P​

Answer 1

Explanation:

I hope it was helpful to you

Answer 2

The correct option regarding final pressure is b) $p× {3}^{ \frac{2}{3} }$.

Given:

A monoatomic gas at a pressure P having a volume v expands isothermally to a volume 3V and then compresses adiabatically to a volume v.

To Find:

The final pressure of the gas.

Solution:

To find the final pressure of the gas we will follow the following steps:

As we know,

The ideal gas equation is

$pv = nrt$

So,

In the isothermal process,

PV = constant which means:

$p1v1 = p2v2$

In the adiabatic process:

$p {v}^{γ} = constant$

Which means:

$p2 {v2}^{γ} =p 3{v3}^{γ}$

According to the question:

P1 = P

P2 =?

P3 =?

V1 = V

V2 = 3V

V3 = V

We have to use the value of the final pressure of the isothermal process because this pressure is the initial pressure for the adiabatic process.

Also,

The value of gamma for monoatomic gas is

$\frac{5}{3}$

Now,

Putting values in the equation of the isothermal process we get,

$pv = p2 \times 3v$

$p2 = \frac{p}{3}$

Now,

Putting values in the equation of the adiabatic process we get,

$\frac{p}{3} {(3v)}^{γ} =p3 \times \: {v}^{γ}$

$\frac{p}{3} \times {3}^{ \frac{5}{3} } \times {v}^{γ} = {v}^{γ} \times p3$

Canceling, volume term both sides we get,

$p3 = \frac{p}{3} \times \sqrt[3]{243} = \frac{p×6.24}{3} = 2.08p = 2p$

$p3 = \frac{p}{3} \times {3}^{ \frac{5}{3} } \: or \: p× {3}^{ \frac{5}{3} -1} = p× {3}^{ \frac{2}{3} }$

Henceforth, the correct option regarding final pressure is b) $p× {3}^{ \frac{2}{3} }$.

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