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When 2 moles of helium gas expand at a constant pressure p= 1.0×10^5 pascals, the temperature increase from 2 °c to 112 °c. 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 exands

Respuesta :

Answer:

1,900 J

Explanation:

The number of moles of helium gas, n = 2 moles

The pressure of the helium gas, p = 1.0 × 10⁔ Pa

The initial temperature of the gas, T₁ = 2°C = 275.15 K

The final temperature of the gas, T₂ = 112°C = 385.15 K

The initial volume of the gas, V₁ = 45 liters

Cp = 20.8 J/(mol·K), Cv = 12.6 J/(mol·K)

The work done by the gas having constant pressure expansion is given as follows;

From the ideal gas law, we have;

[tex]V_2 = \dfrac{T_2 \times n \times R}{P}[/tex]

Where;

R = The universal gas constant = 8.314 J/(mol×K)

Therefore, we get;

[tex]V_2 = \dfrac{385.15 \, K \times 2 \, moles \times 8.314 \, \dfrac{J}{mol \cdot K} }{1.0 \times 10^5 \ Pa} \approx 64.0 \, L[/tex]

The work done, W = P ×ΔV = P × (V₂ - V₁)

∎ W = 1.0 × 10⁔ Pa × (64.0 L - 45.0 L) = 1,900 J

The work done by the gas as it expands, W = 1,900 J.

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