# Container A holds 757 mL of ideal gas at 2.60 atm. Container B holds 164 mL of ideal gas at 4.80 atm. If the gases are allowed to mix together, what is the resulting pressure?

Nov 5, 2016

The ideal gas equation of state is

$P V = n R T$

In this problem we may consider the temperature to be constant, so we can set up an equation that represents the addition of the number of moles of gas in each container (multiplied through by RT).

$\left({P}_{A}\right) \left({V}_{A}\right) + \left({P}_{B}\right) \left({V}_{B}\right) = \left({P}_{f}\right) \left({V}_{f}\right)$

$\left(2.6\right) \left(757\right) + \left(4.8\right) \left(164\right) = \left({P}_{f}\right) \left(757 + 164\right)$

${P}_{f} = 2.99 a t m$

As a check we could assume that $T = 300 K$ and calculate the moles explicitly using $n = \frac{P V}{R T}$:
${n}_{A} = 0.0800 m o l$
${n}_{B} = 0.0320 m o l$
${P}_{f} = \frac{n R T}{V} = \frac{0.112 \times 0.082054 \times 300}{0.921} = 2.99 a t m$