Figure shows two rigid vessels A and B, each of volume `200 cm ^(3)` containg an ideal gas `(C_v = 12.5 JK ^(-1) mol^(-1))`. The vessels are connected to a manometer tube containing mercury. The pressure in both the vessels is 75 cm mercury and the temperature is 300k. (a) `(5-0 J)` of heat is supplied to the gas in the vessels.(b) `(10 J)` to the gas in the vessels B. Assuming no appreciable transfer of heat from A to B calculate the dufference in the heights of mercury in the two sides of the manometer. Gas constant `(R = 8.3 JK^(-1) mol^(-1))`

Explanation

`V=200 cm^3, Cv=12.5 J//mol-K`,
`T=300K, P = 75cm`.
(a) No. of moles of gases in each vessels,
`n=(PV)/(RT)`
`=(75 xx 13.6 xx 980 xx 200)/(8.3 xx 10^7 xx 300)`
`=0.008`
(b) Heat is supplied to the gas but `dV=0`
So, `dQ=dU`
implies `5=nCvdT`
implies `5=0.008 xx 12.5 xx dT`
implies `dT=50` for (A)
Since `(P)/(T) = (P_A)/(T_A)`
implies `75/300 = (P_A 0.008 xx 12.5)/(5)`
`=12.5` cm of Hg
again `P/T = P_B/T_B` (For container B)
implies `75/300 = (P_B 0.008 xx 12.5)/10`
`P_B = 25` cm of Hg.
Mercury moves by a distance
`P_B-P_A = 25 - 12.5 = 12.5` cm.