The pressure exerted by 6.0 g of methane gas in a 0.03 m³ vessel at 129°C is (Atomic masses: C = 12.01, H = 1.01 and R = 8.314 J K⁻¹ mol⁻¹)

Question

Dr. Abubacker · Accepted Answer

[Solution from source] n = 6.0/16.05 = 0.374 mol and T = 129 + 273 = 402 K. P = nRT/V = 0.374 × 8.314 × 402/0.03 ≈ 4.16 × 10⁴ Pa = 41648 Pa. Forgetting to convert °C to K gives a much smaller pressure (13409 Pa). Answer (3).

Q. The pressure exerted by 6.0 g of methane gas in a 0.03 m³ vessel at 129°C is (Atomic masses: C = 12.01, H = 1.01 and R = 8.314 J K⁻¹ mol⁻¹)

Explanation

Reviewed by Dr. Abubacker

Want to attempt the full paper?