Question

A monatomic gas is compressed adiabatically to ${\frac{1}{4}}^{th}$ of its original volume, the final pressure of gas in terms of initial pressure P is:

• A. 7.08 P
• B. 8.08 P
• C. 9.08 P
• D. 10.08

Answer 1

The correct option is D 10.08

We have the equation,

$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$

where,

$P_1$ and $V_1$ are initial pressure and volume of gas respectively.

$P_2$ and $V_2$ are final pressure and volume of gas respectively.

Let $P$ and $V$ be the initial pressure and volume of the gas respectively. Now the gas is compressed adiabativcally to ${\frac{1}{4}}^{th}$ of its initial volume.

Then,

$P{V}^{\gamma }=P_2(\frac{V}{4}{)}^{\gamma }$

$P{V}^{\gamma }=\frac{P_2{V}^{\gamma }}{4^{\gamma }}$

or

$P_2=4^{\gamma }P$

For mono-atomic gases, $\gamma =\frac{5}{3}$

Then,

$P_2=4^{5/3}P=10.08P$