Question

One mole of air $(C_v=\frac{5R}{2})$is confined at atmospheric pressure in a cylinder with a piston at 0$°C$. The initial volume occupied of gas is V. After the equivalent of 13200 J of heat is transferred to it, the volume of gas $V_f$ is nearly Litre. (1 atm = ${10}^{5}N/m^2$)

Answer 1

$P=1atm={10}^{5}N/m^2$ $T=0^{0}C=273K$ $V=\frac{nRT}{P}=\frac{1\times 8.3\times 273}{{10}^5}=0.0227m^3=22.7L$
$C_V=\frac{5}{2}R,C_P=\frac{7}{2}R$
Heat transferred $\Delta Q=n{C}_P\Delta T=\frac{n\left(7R\right)}{2}\Delta T=13200J$
Work done $nR\Delta T=(13200\times \frac{2}{7})$
$=P(V_f-V_i)=3771$
$V_f-V_i=3771\times {10}^{-5}=0.0377m^3$ $V_f=V_i+37.7L=$$22.7L+37.7L$
= $60.4L$