1.

The volume occupied by 2.0 mole of `N_2` at 200 K and 8.21 atm pressure, if `(P_CV_C)/(RT_C)=3/8 and (P_rV_r)/T_r=2.4`, isA. 1.8 LB. `(5/8)` LC. 12.8 LD. 3.6 L

Answer» Correct Answer - D
`(P_rV_r)/T_r=P/T_c V_m/V_c.T_c/T=(PV_m)/(RT).(RT_C)/(P_CV_C)`
`(PV_m)/(RT)=2.4xx(3/8)`
So `V_m=(0.0821xx200)/8.21xx0.9=1.82` So Volume of 2 moles =3.6 L


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