One mole of a gas is enclosed in a cylinde in a cyclinder and occupies a volume of `1.5 L` at a pressure 1.5 atm. It is subjected to strong heating due to which temperature of the gas increase according to the relation `T = alpha V^(2)`, where `alpha` is a positive constant and `V` is volume of the gas.
a. Find the work done by air in increasing the volume of gas to `9 L`.
b. Draw the `P - V` diagram of the process.
c. Determine the heat supplied to the gas (assuming `gamma = 1.5)`.

Explanation

`W = int_(V_(1))^(V_(2)) PdV = int_(V_(1))^(V_(2)) (RT)/(V) dV = R int_(V_(1))^(V_(2)) (alpha V^(2))/(V) dV`
`alpha R int_(V_(1))^(V_(2)) VdV = (alphaR)/(2) (V_(2)^(2) - V_(1)^(2))``alpha R int_(V_(1))^(V_(2)) VdV = (alphaR)/(2) (V_(2)^(2) - V_(1)^(2))`
`W = (R )/(2) (T_(2) - T_(1))` (i)
Now, `T_(1) = alpha V_(1)^(2)`
`T_(2) = alpha V_(2)^(2)`
`T_(2) = T_(1) [(V_(2))/(V_(1))]^(2) = 35 T_(1)`
Hence from Eq. (i) we have
`W = (R )/(2) [36 T_(1) - T_(2)] = (35 RT_(1))/(2) = (35)/(2) P_(1) V_(1)`
`= (35)/(2) xx (1.5 xx 10^(-3)) xx (1.2 xx 10^(5)) J = 31 50 J`
b. We know,
`PV = RT = R alpha V^(2)`
`P prop V`
Hence `P -V` graph is a straight line.
c. `Delta U = n V_(V) Delta T = n ((R )/(gamma - 1)) (T_(2) - T_(1))`
`(R )/(0.5) xx 35 T_(1) = 70 RT_(1) = 70 P_(1) V_(1)`
`= 70 xx (1.2 xx 10^(5)) xx (1.5 xx 10^(-3)) J = 12600 J`
`Delta Q = Delta U + W 12600 + 3150 = 15750 J`