Question

A rigid container containing 10gm gas at some pressure and temperature.The gas has been allowed to escape(do not consider any effusion or diffusion)from the computer due to which pressure of the gas becomes half of its initial pressure and temperature become$(2/3{)}^{rd}$ of its initial.The mass of gas (in gms) escaped is:

• A. $7.5$
• B. $1.5$
• C. $2.5$
• D. $3.5$

Answer 1

The correct option is C $2.5$

$PV=\frac{W}{M}RT$
Initially
$PV=\frac{10}{M}RT$......(1)

Finally $\frac{1}{2}PV=\frac{W}{M}R\frac{2}{3}T$......(2).

Divide equation (2) with equation (1).

$\frac{1}{2}=\frac{W}{10}\times \frac{2}{3}$

$W=\frac{10\times 3}{2\times 2}=\frac{30}{4}=7.5g$.

The mass of gas escaped $=10-7.5=2.5g$.