A gas at 300 K occupies 6.50 L at a pressure of 3.50 atm. What will its pressure be at 250 K if its volume is reduced to 4.80 L?

Question

A gas at 300 K occupies 6.50 L at a pressure of 3.50 atm. 

What will its pressure be at 250 K if its volume is reduced to 4.80 L?

Answer 1

The Ideal Gas Law (PV = nRT) shows us that the pressure, volume, and temperature are related. We can rearrange the equation to get PV/T = nR, which means that if the moles of gas and the R constant remain constant, PV/T will also remain constant.

We can use this idea to set up an equation for this particular problem:

(P₁V₁)/T₁ = P₂V₂/T₂

(3.50 atm * 6.50 L)/(300 K) = (P₂ * 4.80 L)/(250 K)

To isolate P₂, we can multiply both sides by 250 K and divide both sides by 4.80 L:

(3.50 atm * 6.50 L * 250 K)/(300 K * 4.80 L) = P₂

This makes algebraic/mathematical sense, but let's make sure the units also check out:

(3.50 atm * 6.50 L * 250 K)/(300 K * 4.80 L) = P₂

Yep! Liters and K both cancel on the left side, leaving us with P₂ in atm, a unit of pressure.

P₂ = (3.50 atm * 6.50 * 250)/(300 * 4.80) = 3.95 atm

Does this answer make physical sense? Well, we lowered the temperature, which should've caused pressure to drop. However, we also decreased the volume, compressing the gas and increasing its pressure. It turns out that the compression was more impactful than the decrease in temperature, so P₂ was slightly higher than P₁.