Question

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

• A. $37L$
• B. $22L$
• C. $60L$
• D. $30L$

Answer 1

The correct option is C $60L$

Given : $C_v=2.5R$ $T_1=0^{o}C=273.15K$ $P=1atm$

Using ideal gas equation $PV=nR{T}_1$

$\therefore$ $1\times V=1\times 0.082\times 273.15$ $⟹$ $V=22.4$ $L$

$C_p=R+C_v=R+2.5R=3.5R$

Heat transferred at constant pressure $Q=+13200$ $J$

$\therefore$ $Q=n{C}_p\Delta T=n\left(3.5R\right)(T_2-T_1)$

$13200=1\times 3.5\times 8.314\times (T_2-273.15)$ $⟹T_2=726.77K$

As the pressure is constant $⟹\frac{V_2}{T_2}=\frac{V}{T_1}$

$\therefore$ $\frac{V_2}{726.77}=\frac{22.4}{273.15}$ $⟹$ $V_2=60$ $L$