Question

0.4 gm of He in a bulb at a temperature of 'T K had a pressure of 'P' atm. When the bulb was immersed in a hotter bath at a temperature 50 K more than the first one, 0.08 gm of gas had to be removed to restore the original pressure. The value of 'T' is :

Answer 1

$\frac{p_1{v}_1}{h_1{T}_1}=\frac{p_2{v}_2}{h_2{T}_2}$ $\frac{0.32}{4}$
$\frac{p\times v}{\frac{0.4}{4}\times T}=\frac{p\times v}{\frac{0.4-0.08}{4}\times (T+50)}$
$0.08(T+50)=0.1T$
$0.08T+4=0.1T$
$0.02T=4$
$T=\frac{4}{0.02}=200K$