Question

If for an ideal gas, the ratio of pressure and volume is constant and is equal to $1atm{L}^{-1},$ the molar heat capacity at constant pressure would be

• A. $\frac{3}{2}R$
• B. $2R$
• C. $\frac{5}{2}R$
• D. zero

Answer 1

Answer: (B)

By definition

$H=E+PV$

${\left(\frac{dH}{dT}\right)}_P={\left(\frac{dE}{dT}\right)}_P++P{\left(\frac{dV}{dT}\right)}_P=C_{p,m}$

For the given ideal gas, we will have $PV=RT$

or $V^2=RT$ or $2V{\left(\frac{dV}{dT}\right)}_P=R$

or ${\left(\frac{dV}{dT}\right)}_P=\frac{R}{2V}$

$E=\frac{3RT}{2}$

or ${\left(\frac{dE}{dT}\right)}_P=\frac{3}{2}R=C_{v,m}$

$C_p,m=\frac{3}{2}R+P\times \frac{R}{2V}=\left(\frac{3}{2}+\frac{1}{2}\right)R=2R$