What pressure is exerted by 0.622 mol of a gas contained in an 11.5 L container at 18 °C?

Question

What pressure is exerted by 0.622 mol of a gas contained in an 11.5 L container at 18 °C?

Answer 1

The ideal gas law can be used to find the pressure.

PV = nRT

P = nRT / V

P = 0.622 mol x 0.08206 L atm / K mol x 291 K / 11.5 L = 1.29 atm

Answer 2

1 mol. of gas at STP exerts a pressure of 1atm at 273K.

0.622 mol. x 22.4L/mol. = 13.9L at 1atm and 273K.

Using the Combined Gas Law. P1V1T2 = P2V2T1.

1atm x 13.9L x 291K = P2 x 11.5L x 273K

New pressure (P2) = (13.9 x 291) / (11.5 x 273).

= 4,054 / 3,140 = 1.3atm.

Another perfectly correct answer thumbs downed.. Will you please give it up. You can't win.

For your edification, here it is again using other methods...

0.622mol x 22.4L/mol = 13.9L at STP. (1atm and 273K).

13.9L / 11.5L   = 1.212 x 1atm = 1.212atm.

Volume decrease = Pressure increase. (Boyle's Law).

291K / 273K = 1.07 x 1.212atm = 1.292atm .

Temperature increase = Pressure increase (g*y Lussac's Law)....(1.3atm is quite acceptable as a negligible difference of 0.008atm).

(Not much chance of an apology, I suppose ???. You know I know who).

Answer 3

P= nRT/V

= 0.622*0.0821*291/11.5 = 1.21