Question

A vessel is filled with an ideal gas at a pressure of $10\text{atm}$ and temperature of $27{}^{\circ }\text{C}$. Half of the mass is removed from the vessel and temperature of the remaining gas is increased to $87{}^{\circ }\text{C}$. Then the pressure of the gas in the vessel will be

• A. $5\text{atm}$
• B. $6\text{atm}$
• C. $7\text{atm}$
• D. $8\text{atm}$

Answer 1

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

Given:
Initial pressure, $P_1=10\text{atm}$
Initial temperature, $T_1=27{}^{\circ }\text{C}=300\text{K}$
Final temperature, $T_2=87{}^{\circ }\text{C}=360\text{K}$
To find:
Final pressure, $P_2=?$

We know that, ideal gas equation is given by,
$PV=nRT$, where $P$ is pressure, $V$ is volume, $n$ is number of moles, $R$ is gas constant and $T$ is temperature of the gas.
So, $n=\frac{PV}{RT}=\frac{10V}{300R}=\frac{V}{30R}$
As, half of the mass is removed.
$\therefore n_2=\frac{n}{2}=\frac{V}{60R}$ .......(1)
Now, $P_2=\frac{n_{2}R{T}_2}{V_2}$
$\Rightarrow P_2=\frac{360RV}{60RV}$
[ from (1) and volume will be same as container is the same ]
$\Rightarrow P_2=6\text{atm}$