Question

Container A holds 767 mL of ideal gas at 2.80 atm. Container B holds 129 mL of ideal gas at 4.40 atm. If the gases are allowed to mix together, what is the resulting pressure?

Answer #1

assuming that the temperature remains same,

Vtotal = 767 + 129 =896 ml

n_{total } =P1V1/RT +
P2V2/RT = (767x2.8 + 129x4.4)/RT

new Total pressure = Pnew =
(n_{Total}RT)/V_{total } =
{(767x2.8 + 129x4.4)/RT} X RT/896 ml

= {(767x2.8 + 129x4.4)} /896 Atm = 3 .0 3 atm

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