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 412 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 412 K at constant pressure by doing work W2 while transferring energy Q2 by heat.

Required:
Find ÎEint, 1, Q1, and W1 for the process at constant volume.

Answer 1

ΔE
`_{int`,₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J

Step-by-step explanation:

Given the data in the question;

T
`_i` = 300 K, T
`_f` = 412 K, n = 0.7 mol

since helium is monoatomic;

Cv = (3/2)R, Cp = (5/2)R

W₁ = 0 J [ at constant volume or ΔV = 0]

Now for the first cylinder; from the first law of thermodynamics;

Q₁ = ΔE
`_{int`,₁ + W₁

Q₁ = ΔE
`_{int`,₁ = n × Cv × ΔT

we substitute

Q₁ = ΔE
`_{int`,₁ = 0.7 × ( 3/2 )8.314 × ( 412 - 300 )

Q₁ = ΔE
`_{int`,₁ = 0.7 × 12.471 × 112

Q₁ = ΔE
`_{int`,₁ = 977.7 J

Therefore, ΔE
`_{int`,₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J