Question

Interested in doing part B

1. a) One mole of an ideal gas is compressed irreversibly from 2 L to 1 L under a constant external pressure of 5 atm. The temperature is 300 K. Calculate the work done on the gas during the compression. b When the gas is cooled to sufficiently low temperatures, it is found experimentally that the equation of state for the gas no longer resembles the ideal gas law. Instead, what is found is that the pressure of the gas obeys the following empirical relation: NT where σ, γ and α are constants. Derive an expression for the work that the gas performs in expanding reversibly and isothermally from some initial to some final volume.

Answer 1

(a) P = 5 atm = 5 times 1.01325 times 10^5 pa V_2 = 1L - 1 times 10^-3 m^3 V_1 = 2L = 2 times 10^-3 m^3 Compressed irreversibly isothermally work done on gas = -P_ext delta V = -p_ext (v_2 - v_1) = -5 times 1.01 times 10^5 (1 - 2) times 10^-3 = + 5 times 1.01 times 10^5 times 1 times 10^-3 TildeEqual 5.06 times 10^2 J Ans (b) pressure of gas now obeys P = a squareroot T/v^3 + Y T^2 e^-av equation (1) irreversibly and isothermally (T = constant) Total amount of work done by isothermal revertible expansion from v_1 to v_2 w = integral^v_2_v_1 p dv = integral^v_2_v_1 (a squareroot T/v^3 + Y T^2 e^-av) dv = [a squareroot T v^-3 + 1/-3 + 1 + YT^2 e^-av/-a]^v_2_v_1 [as T = const] = [-1/2 a squareroot T v^-2 YT^2/a e^-av]^v_2_v_1 = 1/2 a squareroot T [1/v_1^2 - 1/v_2^2] - YT^2/a [e^-av_2 - e^-av_1] Ans