Question

Two identical cylinders with a movable piston contain 0.7 mol of helium gas at a temperature of 300 K. The temperature of the gas in the first cylinder is increased to 564 K at constant volume by doing work W1 and transferring energy Q1 by heat. The temperature of the gas in the second cylinder is increased to 564 K at constant pressure by doing work W2 while transferring energy Q2 by heat.
A. Find ΔEint, 1, Q1, and W1 for the process at constant volume.
B. Find ΔEint, 2, Q2, and W2 for the process at constant pressure.

Answer 1

Answer:

Explanation:

A ) At constant volume :

ΔEint = n Cv x ΔT , n is no of moles , Cv is specific heat at constant volume , ΔT is increase in temperature .

For helium Cv = 3/2 R = 1.5 x 8.3 J = 12.45 J

ΔEint = .7 x 12.45 x ( 564 - 300 )

= 2300.76 J .

W₁ = 0 because volume is constant so work done by gas is zero .

Q₁ = ΔEint = 2300.76 J

B )

At constant pressure

Q₂ = n Cp Δ T , Cp is specific heat at constant pressure .

For monoatomic gas ,

Cp = 5/2 R = 2.5 x 8.3 J = 20.75 J

Q₂ = .7 x 20.75 x 264 J

= 3834.6 J

W₂ = work done by gas

= PΔV = nRΔT

= .8  x 8.3 x 264

= 1752.96 J

ΔEint = Q₂ - W₂

= 3834.6 - 1752.96

= 2081.64 J.