Answer:
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;
![V = \dfrac{n \cdot R \cdot T}{P}](https://tex.z-dn.net/?f=V%20%3D%20%5Cdfrac%7Bn%20%5Ccdot%20R%20%20%5Ccdot%20T%7D%7BP%7D)
Plugging in the values gives;
![V_2 = \dfrac{n \cdot R \cdot T_2}{P}](https://tex.z-dn.net/?f=V_2%20%3D%20%5Cdfrac%7Bn%20%5Ccdot%20R%20%20%5Ccdot%20T_2%7D%7BP%7D)
Where;
V₂ = The volume of the gas at state 2
Therefore;
![V_2 = \dfrac{2 \cdot 8.314 \cdot 385.15}{1.0 \times 10^5} \approx 0.064](https://tex.z-dn.net/?f=V_2%20%3D%20%5Cdfrac%7B2%20%5Ccdot%208.314%20%20%5Ccdot%20385.15%7D%7B1.0%20%5Ctimes%2010%5E5%7D%20%5Capprox%200.064)
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