A cylinder with a volume of V = 50 l contains compressed air under a pressure of 1 P = 50 atm.

| April 3, 2021 | education

A cylinder with a volume of V = 50 l contains compressed air under a pressure of 1 P = 50 atm. After part of the air was discharged outside, the pressure in the cylinder dropped to 2 P = 30 atm. How much did the air mass in the cylinder decrease? Consider that the air temperature in the cylinder remains constant and equal to t = 22C. Molar mass of air M = 29 g / mol.

To determine the decrease in the mass of air in a given cylinder, we apply the formula: Pk * V = mk * R * T / MHe; Pк = (mн – Δm) * R * Т / (V * Мair) = Рн – Δm * R * Т / (V * Мair), whence we express: Δm = (Рн – Рк) * V * Мair / (R * T).

Constants and variables: Pн – initial air pressure (Pн = 50 atm = 5 * 10 ^ 6 Pa); Рк – steady-state pressure (Рк = 30 atm = 30 * 10 ^ 6 Pa); V is the volume of the cylinder (V = 50 l = 0.05 m3); Mair – air mass (according to the condition Mair = 29 g / mol = 0.029 kg / mol); R – universal gas constant (R = 8.31 J / (K * mol)); Т – constant air temperature (Т = 295 К (22 ºС)).

Let’s calculate: Δm = (Рн – Рк) * V * Мair / (R * Т) = (5 * 10 ^ 6 – 3 * 10 ^ 6) * 0.05 * 0.029 / (8.31 * 295) = 1.183 kg.

Answer: The air mass in this cylinder has decreased by 1.183 kg.



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