Question

An open flask has Helium gas at 2atm and 327${}^{\circ }C$. The flask is heated at 527${}^{\circ }C$ at the same pressure. The fraction of original gas remaining in the flask is?

• A. $\frac{3}{4}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $\frac{2}{5}$

Answer 1

The correct option is A $\frac{3}{4}$

Initial pressure, $P_1=1atm$

Initial temperature, $T_1={327}^{o}C=600K$

Final pressure, $P_2=1atm$

Final temperature, $T_2={527}^{o}C=800K$

now, $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

$\Rightarrow V_2=\frac{1\times V_1}{600}\times \frac{800}{1}=\frac{4}{3}{V}_1$

But the flask can hold only $V_1$ volume of air,

$\therefore$ Volume of air expelled= $\frac{4}{3}{V}_1-V_1=\frac{V_1}{3}$

$\therefore$ Fraction of air expelled out= $\left(\frac{V_1}{3}\right)/\left(4V_1/3\right)$

$\therefore$ Fraction of air remaining in flask= $1-\frac{1}{4}=3/4$