Two rigid adiabatic vessels A and B which initially, contain two gases at different temperature are connected by pipe line with value of negligible volume. The vessel 'A' contain 2 moles Ne gas `(C_(p.m) = (5)/(2)R)` at 300K, vessel 'B' contain 3 moles of `SO_(2)` gas `(C_(p.m) = 4R)` at 400K. The volume of A & B vessel is 4 and 6 litre respectively. The final total pressure (in atm) when value is opened and 12 Kcal heat supplied through it to vessels. [Use: R= 2 cal/mol, K and R= 0.08L. atm/mol K as per desire]

Options

A. 3.5 atm
B. 7 atm
C. 35 atm
D. 70 atm

Answer

C

Explanation

This total process was carried out at constant volume. So we have to use speciphic heat at constant volume. So after the opening of the value initially both gases attain same temperature T as per the following equation
`2 (3)/(2) R (T - 300) = 3 xx 3R (400-T)`
On solving T= 375
Again temperature will be increases to `T^(1)` due to the supply of 12K. Cal of heat as follows:
`2 xx (3)/(2) R (T^(1) - 375) + 3 xx 3R (T^(1) - 375) = 12000`
On solving `T^(1) = 875`
So, `P = (5 xx 0.08 xx 875)/(10) = 35` atm