Question

Calculate the pressure, in atmospheres, of 2.40 moles helium gas in a 8.00 L container at 18 °C.

Answer 1

According to the ideal gas law, $pV=nRT,$ where p is the pressure, V is the volume, n is amount in moles, T is the temperature in Kelvin, and R is a gas constant that equals $0.082057 L\cdot{atm\cdot{mol^{-1}\cdot{K^{-1}}}}$.

$T = 18+273.15=291.15K$

Therefore,

$p=\frac{nRT}{V}=\frac{2.40mol\times0.082057 L\cdot{atm\cdot{mol^{-1}\cdot{K^{-1}}}}\times{291.15K}}{8.00L}=7.17atm$

Answer: 7.17 atm