A rigid and insulated tank of `3m^(3)` volume is divided into two compartments. One second compartment of volume of `2m^(3)` contains an ideal gas at 0.8314 Mpa and 400K and while the second compartment of volume `1m^(3)` contains the same gas at 8.314 MPa and 500K. If the partition between the two compartments isruptured. the final temperature of the gas is:

Options

A. 420K
B. 450K
C. 480K
D. none of these

Answer

C

Explanation

Number of moles in the first compartment = `(PV)/(RT) = (8.314 xx 10^(5) xx 2)/(8.314 xx 400) = 500`. Number of moles in the second compartment `=(8.314 xx 10^(6) xx 1)/(8.314 xx 500) = 2000`
Heat lost by the gas in second compartment = Heat gained by the gas in first compartment
So `2000 xx (500-T) xx C = 500 xx (T - 400)xx C`
On solving, T= 480K