if i have 7.7 moles of a gas at a pressure of 0.09 ATM and a temperature of 329 K, what is the volume of the container that the gas is in?

N=number of moles=7.7 moles.

P=pressure at atm=0.09 atm

T=temperature at ºK=329 ºK

R=universal gas costant=0.082 (L*atm) / (ºK*mol)

The ideal gas law:

PV=nRT

therefore:

0.09 atm*V=7.7 moles * 0.082 (L*atn) / (ºK * mol) * 329ºK

V = (7.7 moles * 0.082 (L*atn) / (ºK * mol) * 329ºK) / (0.09 atm) = 2,308.12 L

Answer: the volume is 2,308.12 litres