The density of a polyatomic gas in standard conditions is $0.795\,Kg/{m^3}$ . The specific heat of the gas at constant volume is:
A) $930\,JK{g^{ - 1}}{K^{ - 1}}$
B) $1400\,JK{g^{ - 1}}{K^{ - 1}}$
C) $1120\,JK{g^{ - 1}}{K^{ - 1}}$
D) $1600\,JK{g^{ - 1}}{K^{ - 1}}$

Hint: Substitute the known values and the constant in the formula of the ideal gas equation to calculate the unknown mass of the polyatomic gas. Substitute the mass value in the second formula to calculate the specific heat capacity of the gas.

Useful formula:
(1) The ideal gas equation is
$PM = dRT$
Where $P$ is the pressure of the gas, $M$ is the molar mass, $d$ is the density of the gas, $R$ is the universal gas constant and $T$ is the temperature.

(2) The specific heat capacity is given as
$s = \dfrac{f}{2}R \times \dfrac{{1000}}{M}$
Where $s$ is the specific heat capacity and $f$ is the degrees of freedom.

Complete step by step solution:
It is given that the
Density of the polyatomic gas, $d = 0.795\,Kg{m^{ - 3}}$
Use the formula(1) of the ideal gas equation,
$PM = dRT$
Substitute that the pressure is equal to $1\,atm$ , density is given, the universal gas constant is
$8.314 J{mol}^{ - 1}{K^{ - 1}}$ and the temperature as $273\,K$ in the above formula.
$1 \times M = 0.795 \times 8.314 \times 273$
By simplifying the above step, we get
$M = 18\,Kg$
Substitute the value of the mass in the formula of the specific heat capacity,
$s = \dfrac{f}{2}R \times \dfrac{{1000}}{M}$
The degrees of freedom is $6$ and hence substituting the known parameters in the above step.
$s = \dfrac{6}{2} \times 8.314 \times \dfrac{{1000}}{{18}}$
By simplification of the above step, we get
$s = 1400\,JK{g^{ - 1}}{K^{ - 1}}$
Hence the value of the specific heat capacity is obtained as $1400\,JK{g^{ - 1}}{K^{ - 1}}$ .

Thus the option (B) is correct.

Note: The mass value from the ideal gas equation is calculated as $18$ . From this value, it is clearly understood that the molecule is water. That is why the degrees of freedom is taken as $6$ and it gets substituted in the formula to calculate the specific heat of the water.