Question

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

• A. $5$ atm
• B. $6$ atm
• C. $7$ atm
• D. $8$ atm

Answer 1

Answer: (B)

$6$ atm

Given : Initial temperature of the gas $T_i={27}^{o}C=300K$

Initial pressure of the gas $P_i=10$ atm

Final temperature of the gas $T_f={87}^{o}C=360K$

Final mole of gas in the vessel $n_f=\frac{n_i}{2}$

Using ideal gas equation, $PV=nRT$

Initially, $10V=n_{i}R\left(300\right)$ ..........(1)

Let the final pressure in the vessel be $P_f$.

Finally, $P_{f}V=\frac{n_i}{2}R\left(360\right)$ ...........(2)

Dividing (1) and (2), we get $\frac{P_f}{10}=\frac{360}{2\times 300}$ $⟹P_f=6$ atm