`5.6dm^(3)` of an unknown gas at `S.T.P` requires 52.25J of heat to raise its temperature by `10^(@)C` at constant volume. Calculate `C_(V),C_(P)` and atomicity of the gas

Explanation

`22.4dm(3)` of a gas at S.T.P. `=1` mol ` :. 5.6 dm^(3)` of the gas at S.T.P. `= (1)/(22.4) xx 5.6 =0.25 mol`
Thus, for `10^(@)` rise , `0.25` mol of the gas at constant volume require heat `= 52.25 J`
`:. `For `1^(@)` rise, 1 mol of the gas at constant volume will require heat `= (52.25)/(10 xx 0.25) J = 20.9 J`
`:. C_(v) = 20.9 J K^(-1) mol^(-1)`
Now, `C_(p) = C_(v) + R = 20. 9 J K^(-1) mol^(-1) + 8. 314 J K^(1) mol^(-1) = 29.214 J K^(-1) mol^(-1)`
`:. gamma = (C_(p))/( C_(v))= (29.214)/(20.9) = 1.4`. Hence, the gas is diatomic