The molar specific heat at constant pressure of an ideal gas is $$\left( {\frac{7}{2}} \right)R.$$ The ratio of specific heat at constant pressure to that at constant volume is
A.
$$\frac{7}{5}$$
B.
$$\frac{8}{7}$$
C.
$$\frac{5}{7}$$
D.
$$\frac{9}{7}$$
Answer :
$$\frac{7}{5}$$
Solution :
We have given molar specific heat at instant pressure $${C_p} = \frac{7}{2}R$$
Mayer's relation can be written as :
Molar specific heat at constant pressure - Molar specific heat at constant volume = Gas constant,
$$\eqalign{
& {\text{i}}{\text{.e}}{\text{.}}\,{C_p} - {C_V} = R \Rightarrow {C_V} = {C_p} - R \cr
& = \frac{7}{2}R - R = \frac{5}{2}R\left[ {\because {C_p} = \frac{7}{2}R} \right] \cr} $$
Hence, required ratio is $$\gamma = \frac{{{C_p}}}{{{C_V}}} = \frac{{\left( {\frac{7}{2}} \right)R}}{{\left( {\frac{5}{2}} \right)R}} = \frac{7}{5}$$
Releted MCQ Question on Heat and Thermodynamics >> Thermodynamics
Releted Question 1
An ideal monatomic gas is taken round the cycle $$ABCDA$$ as shown in the $$P - V$$ diagram (see Fig.). The work done during the cycle is
If one mole of a monatomic gas $$\left( {\gamma = \frac{5}{3}} \right)$$ is mixed with one mole of a diatomic gas $$\left( {\gamma = \frac{7}{5}} \right)$$ the value of $$\gamma $$ for mixture is
A closed compartment containing gas is moving with some acceleration in horizontal direction. Neglect effect of gravity. Then the pressure in the compartment is
A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature $$T.$$ Neglecting all vibrational modes, the total internal energy of the system is