Answer to Question #75600 in Physical Chemistry for Jyoti chaudhary

Question

Calculate the work done when 1.00 mil of an ideal gas expands isothermally and reversibly at 3.00*10^2 K from 2.00* 10^5 Pa

Answer 1

To solve the problem we use the formula:
W = -2.303*n*R*T*log(P1/P2)
How did pressure change in this task? … from 2.00 *106 Pa to 2.00 *105 Pa? If this is the case then the problem is solved by the following way:
n = 1 , R = 8.314 J/(K*mol) , T = 3 * 10² K;
P₁ = 2 * 10⁶ Pa , P₂ = 2 * 10⁵ Pa;
W = -2.303 *1*8.314*3 * 10² * log(2 * 10⁶/2 * 10⁵)
W = -5744.1426 J = -5.744 kJ.
Answer: W = -5.744 kJ