Question

An open beaker at ${27}^o$ is heated to ${477}^o$ what fraction of air would have been expelled out?

Answer 1

suppose at 27${}^{o}C$ that is 300 k the volume is v.

but 477${}^{o}C$ that is 750 k the volume is xv.

using the gas law

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

$V_2=V_1\times \frac{T_2}{T_1}$

$V_2=V_1\times \frac{750}{300}$

If $V_1=V$

so the final volume is 2.5 v

so the volume expelled out = 2.5 - 1 = 1.5 v