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. 420 K
B. 450 K
C. 480 K
D. None of these

Answer

C

Explanation

Mole of the gas in the first compartment
`n_(1)=(P_(1)V_(1))/(RT_(1))=-(0.8314xx10^(6)xx2)/(8.314xx400)=500`
Similarly, `n_(2)=2000`
The tank is rigid and insulated hence w = 0
and q = 0 therefore `DeltaU=0`
Let `T_(f)` and `P_(f)` denote the final temperature and pressure respectively
`DeltaU=n_(1)C_(V,m)[T_(f)-T_(1)]+n_(2)C_(V,m)[T_(f)-T_(2)] =0`
`500(T_(f)-400)+2000(T_(f)-500)=0`
`T_(f)=480K`