One mole of air `(C_(V) = 5R//2)` is confined at atmospheric pressure in a cylinder with a piston at `0^(@)C`. The initial volume occupied by gas is`V`. After the equivalent of `13200 J` of heat is transferred to it, the volume of gas `V` is nearly `(1 atm = 10^(5) N//m^(3))`

Options

A. `37 L`
B. `22 L`
C. `60 L`
D. `30 L`

Answer

C

Explanation

c. `P = 1 atm = 10^(5) N//m^(2)`
`T = 0^(0) C = 273 K`
`V = (nRT)/(P) = (1 xx 8.3 xx 273)/(10^(5)) = 0.0227 m^(3) = 22.7 L`
`C_(V) = (5)/(2) R, C_(P) = (7)/(2) R`
Heat transferred
`Delta Q = nC_(P) Delta T = n(7R)/(2) Delta T = 13200 J`
Work done
`nR Delta T = (13200 xx 2)/(7)`
`= P (V_(f) - V_(i)) = 3771`
`V_(f) - V_(i) = 3771 xx 10^(-5) = 0.0377 m^(3)`
`V_(f) = V_(i) + 37.7 L`
`22.7 L + 33.7 L`
`60.4 L ~~ 60 L`