Question

2.8 × 10^-2 kg of nitrogen is expanded isothermally and reversibly at 300 K from 15.15 × 10⁵ Pa when the work done is found to be -17.33 kJ. Find the final pressure.

Answer 1

Given : 

Mass of nitrogen = m = 2.8 × 10^-2 kg

Temperature = T = 300 K

Work obtained in expansion = W_max = -17.33 kJ

= – 17330 J

Initial pressure = P₁ = 15.15 × 10⁵ Pa

= 1.515 × 10⁶ Pa

Molar mass of nitrogen (N₂) = MN₂

= 28 × 10^-3 kg mol^-1

Final pressure = P₂ = ?

Number of moles of N₂ = n = \(\frac{m}{MN_2}\)

= \(\frac{2.8\times 10^{-2}}{28\times 10^{-3}}\) = 1 mol

W_max = -2.303 × nRT log₁₀\(\frac{P_1}{P_2}\)

17330 = 2.303 × 1 × 8.314 × 300 × log₁₀\(\frac{1.515 \times 10^6}{P_2}\)

∴ \(\frac{17330}{2.303\times 1\times 8.314 \times 300}\)

= [log₁₀1.515 × 10⁶ – log₁₀P₂]

3.017 = 6.1804 – log₁₀P₂

∴ log₁₀P₂ = 6.1804 – 3.017 = 3.1634

∴ P₂ = Antilog 3.1634

= 1456.8 Pa

∴ Final pressure = 1456.8 Pa