At a given volume and temperature, the pressure of a gas ……….. .

Explanation

`Cv = Cp - R = 2.5 R - R = 1.5 R`.
The amount of the gas (in moles) is `n pV / RT
`(1.6 xx 10^(6)N m^(-2) xx (0.0083 m^(3)) / (8.3 J K^(-1) mol^(-1)) (300k) = 5.3mol`.
As the gas is kept in a closed vessel, its volume is constant. Thus, we have
`DeltaQ = n Cv DeltaT
or, DeltaT = DeltaQ /nCv`
= `2.49 xx 10^(4) J /(5.3 mol) (1.5 xx 8.3 J K^(-1) mol^(-1) = 377 K`.
The final temperature is `300 K + 377 K = 677 K`.
We have ,
`P^(1)V^(1) / T^(1) = p^(2)V^(2) / T^(2)`.
Here `V^(1) = V^(2) Thus,
P^(2) = T^(2) / T^(1) P^(1) = 677 / 300 xx 1.6xx 10^(6) N m^(-2)
=3.6 xx 10^(6) N m^(-2).