A rigid cylinder of volume 25 ~L contains 140 ~g of nitrogen gas ( N₂ ) at 27^ C . A valve is opened and 56 ~g of the gas is allowed to leak out. If the temperature of the cylinder is allowed to return to 27^ C , what is the final gauge pressure of the gas in the cylinder? [Given: R = 25 3 ~J ~mol ⁻¹ ~K ⁻¹ , 1 ~atm = 10^5 ~N/m ^2 , and molar mass of N₂ = 28 ~g/mol ]

Step-by-step solution

Initial mass of N₂ gas = 140 ~g Mass of gas leaked = 56 ~g Final mass of gas remaining in the cylinder = 140 - 56 = 84 ~g Number of moles remaining, n = 84 28 = 3 ~mol Using the ideal gas equation for the final state: P_ abs V = nRT P_ abs = nRT V = 3 ( 25 3 ) 300 25 10⁻³ = 25 300 25 10⁻³ = 3 10^5 ~Pa Since 1 ~atm = 10^5 ~Pa , the absolute pressure is 3 ~atm . Gauge pressure is the pressure in excess of atmospheric pressure: P_ gauge = P_ abs - P_ atm = 3 ~atm - 1 ~atm = 2 ~atm Answer: 2 ~atm