. When 2 moles of helium gas expand at constant pressure P = 1.0 × 105 Pascals, the temperature increases from 2℃ to 112℃. If the initial volume of the gas was 45 liters. Cp=20.8J/mol. K, Cv=12.6J/mol. K. Determine i. The work done W by the gas as it expands (4) ii. The total heat applied to the gas (2) iii. The change in internal energy (2

i. Work done by the gas as it expands is approximately 1,900 J

ii. The total heat supplied is approximately 4, 576 J

iii. The change in internal energy is approximately 2,772 J

Explanation:

The constant pressure of the helium gas, P = 1.0 × 10⁵ Pa

The initial and final pressure of the gas, T₁, and T₂ = 2°C (275.15 K) and 112°C (385.15 K) respectively

The number of moles of helium in the sample of helium gas, n = 2 moles

The volume occupied by the gas at state 1, V₁ = 45 L

i. By ideal gas law, we have;

P·V = n·R·T

Therefore;

Plugging in the values gives;

Where;

V₂ = The volume of the gas at state 2

Therefore;

The volume of the gas at state 2, V₂ ≈ 0.064 m³ = 64 Liters

Work done by the gas as it expands, W = P × (V₂ - V₁)

∴ W ≈ 1.0 × 10⁵ Pa × (64 L - 45 L) = 1,900 J

Work done by the gas as it expands, W ≈ 1,900 J

ii. The total heat supplied, Q = Cp·n·ΔT

∴ Q = 20.8 J/(mol·K) × 2 moles × (385.15 K - 275.15 K) = 4,576 J

The total heat supplied, Q = 4, 576 J

iii. The change in internal energy, ΔU = Cv·n·ΔT

∴ ΔU = 12.6 J/(mol·K) × 2 moles × (385.15 K - 275.15 K) = 2,772 J

The change in internal energy, ΔU = 2,772 J