Respuesta :
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;
[tex]V = \dfrac{n \cdot R \cdot T}{P}[/tex]
Plugging in the values gives;
[tex]V_2 = \dfrac{n \cdot R \cdot T_2}{P}[/tex]
Where;
V₂ = The volume of the gas at state 2
Therefore;
[tex]V_2 = \dfrac{2 \cdot 8.314 \cdot 385.15}{1.0 \times 10^5} \approx 0.064[/tex]
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
