Question

Air is filled at ${60}^o$C in a vessel of open mouth. The vessel is heated at temperature T so that $\frac{1}{4}$ th part of air escapes. Assuming volume of container remaining constant, find value of T.

• A. ${80}^{o}C$
• B. ${444}^{o}C$
• C. ${333}^{o}C$
• D. ${171}^{o}C$

Answer 1

Answer: (D)

${171}^{o}C$

d.$M_1=M,T_1=60+273=333K$
$M_2=M-\frac{M}{4}=\frac{3M}{4}$
(as $1/4$ th part of air escapes If pressure and volume of the gas remain constant, then $MT=$ constant
$\therefore \frac{T_2}{T_1}=\frac{M_1}{M_2}=\left(\frac{M}{3M/4}\right)=\frac{4}{3}$
$\Rightarrow T_2=\frac{4}{3}\times T_1=\frac{4}{3}\times 333=444K={171}^{\circ }C$