Pls solve the no 2 sum its urgent.

Pls solve the no 2 sum its urgent. 0 00

22.4 dm³ of gas at S.T.P = 1 Mole
Therefore

5.6 dm³ of gas  = $\frac{1}{22.4}\times 5.6 = 0.25 moles$


For temperature rise in 10 ^oC ,0.25 mole of gas at constant volume requires = 52.25 J

For rise in 1 ^oC , 1 mole of gas at constant volume will require  = $\frac{52.25}{10\times 0.25}=20.9$ J K^-1 mol^-1

Therefore
C_v = 20.9 J K^-1 mol^-1

We know C_P - C_v = R

 C_P ​​= R + C_v = 8.314 J K^-1 mol^-1 + 20.9 J K^-1 mol^-1 =29.214 J K^-1 mol^-1

​​C_P ​​= 29.214 J K^-1 mol^-1


2)
​ $$\begin{gathered} \gamma =\frac{C_P}{C_v} \\ \gamma =\frac{29.14}{20.9} = 1.39 \approx 1.4 \\ Hence the gas is diatomic. \end{gathered}$$