Question
One mole of a gas occupies $$22.4\,lit$$ at $$N.T.P.$$ Calculate the difference between two molar specific heats of the gas. $$J = 4200\,J/kcal.$$
A.
$$1.979\,k\,cal/kmol\,K$$
B.
$$2.378\,k\,cal/kmol\,K$$
C.
$$4.569\,k\,cal/kmol\,K$$
D.
$$3.028\,k\,cal/kmol\,K$$
Answer :
$$1.979\,k\,cal/kmol\,K$$
Solution :
$$V = 22.4\,litre = 22.4 \times {10^{ - 3}}{m^3},J = 4200\,J/kcal$$
by ideal gas equation for one mole of a gas,
$$\eqalign{
& R = \frac{{PV}}{T} = \frac{{1.013 \times {{10}^5} \times 22.4 \times {{10}^{ - 3}}}}{{273}} \cr
& {C_p} - {C_v} = \frac{R}{J} = \frac{{1.013 \times {{10}^5} \times 22.4}}{{273 \times 4200}} = 1.979\,kcal/kmol\,K \cr} $$