Question

A sample of 80.0 mL of krypton gas, Kr, is at 756 torr and 18.5°C. What is the new volume if the pressure is decreased to 185 torr and the temperature is decreased to 11.0°C?

Answer 1

Initi al pressure, P = 756 torr Initi al volume, Vi = 80.0 mL Initi al temperature, 77 = 18.5°C = 18.5+273= 291.5K final pressure, P, = 185 torr final volume, V = ? final temperature, T, = 11.0°C= 11.0+ 273= 284 K For ideal gas equation. Substitute the corresponding values in the above relation and calculate v 7,22 756 torr 80.0 mL x 284K 291.5 Kx185 torr = 318.5 mL Therefore, the required solution is 318.5 mL - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (2) Idel a gas equation is PV = nRT PV = RT - MPV .... (1) where, misthe mass of nitrogen gas. m= RT Temperature of nitrogen, T = 105 °C= (105+273)K = 378 K Volume, V = 5.00L pressure, P = 0.92 atm gas constant, R= 0.08206 L atm/molK Molar mass of Nitrogen, M = 28.0 g/mol Substitute the corresponding values in the above relation and calculate m m = MPV m= RT 28.0 g/mol x 0.92 atm x 5.00 L m = 0.08206 L atm/molKx378 K m= 4.15 g Therefore, the required solution is 4.15 g