Question

A gas is present at a pressure of $2\text{atm}$. What should be the increase in pressure so that the volume of the gas can be decreased to $\frac{1}{4}\text{th}$ of the initial value, if the temperature is maintained constant?

• A. $8\text{atm}$
• B. $2\text{atm}$
• C. $6\text{atm}$
• D. $10\text{atm}$

Answer 1

The correct option is C $6\text{atm}$

If temperature is maintained constant, according to Boyle's law, we can say that
$PV$ = Constant, for a given mass where $P$ is pressure and $V$ is volume.
$\Rightarrow P_1{V}_1=P_2{V}_2$
Given,
$P_1=2\text{atm}$
$V_1=V{\text{m}}^3$
$V_2=\frac{V}{4}{\text{m}}^3$

i.e $2\times V=P_2\times \frac{V}{4}$
$\Rightarrow P_2=8\text{atm}$
Initial pressure is $P_1=2\text{atm}$ and final pressure is $P_2=8\text{atm}$
$\therefore$ Increase in pressure $\Delta P=P_2-P_1=8-2=6\text{atm}$
Thus, option (c) is the correct answer.