Combined Gas Law Practice

Question

2) I have an unknown volume of gas at a pressure of 0.80 atm and a temperature of 328 K. If I raise the pressure to 1.5 atm, decrease the temperature to 323 K, and measure the final volume to be 50 liters, what was the initial volume of the gas?

3) If I have 20 liters of gas held at a pressure of 78 atm and a temperature of 905.0 K, what will be the volume of the gas if I decrease the pressure to 50 atm and decrease the temperature to 750.0 K?

4) If I have 3.0 L of gas at a pressure of 5.1 atm and a temperature of 52.0 0C, what will be the temperature of the gas if I decrease the volume of the gas to 2.6 L and decrease the pressure to 3.2 atm?

5) I have an unknown volume of gas held at a temperature of 110 K in a container with a pressure of 62 atm. If by increasing the temperature to 225 K and decreasing the pressure to 33 atm causes the volume of the gas to be 29 liters, how many liters of gas did I start with?

Answer 1

Answers: 2) 95 L · 3) 26 L · 4) 177 K (−96 °C) · 5) 7.5 L

The combined gas law

P₁V₁/T₁ = P₂V₂/T₂

It contains Boyle's, Charles's, and Gay-Lussac's laws as special cases — hold any one variable constant and it collapses to the familiar two-variable form. Rather than memorising rearrangements, write the equation, put the unknown on one side, and multiply across.

2. P₁ = 0.80 atm, T₁ = 328 K; P₂ = 1.5 atm, T₂ = 323 K, V₂ = 50 L. Find V₁.

V₁ = (P₂V₂T₁) / (P₁T₂) = (1.5)(50)(328) / [(0.80)(323)] = 24 600 / 258.4 = 95 L

Check: the pressure was nearly doubled while the temperature barely moved, so the volume should have roughly halved — 95 L down to 50 L. Consistent.

3. V₁ = 20 L, P₁ = 78 atm, T₁ = 905.0 K; P₂ = 50 atm, T₂ = 750.0 K. Find V₂.

V₂ = (P₁V₁T₂) / (P₂T₁) = (78)(20)(750) / [(50)(905)] = 1 170 000 / 45 250 = 26 L

Check: two competing effects — dropping the pressure expands the gas, cooling it shrinks the gas. The pressure change is the larger factor (78/50 = 1.56 versus 905/750 = 1.21), so expansion wins and the volume rises modestly. It does.

4. V₁ = 3.0 L, P₁ = 5.1 atm, T₁ = 52.0 °C = 325.15 K; V₂ = 2.6 L, P₂ = 3.2 atm. Find T₂.

T₂ = T₁ × (P₂V₂) / (P₁V₁) = 325.15 × (3.2)(2.6) / [(5.1)(3.0)] = 325.15 × 0.5438 = 177 K, i.e. −96 °C

This is the only question in the set given in Celsius, and it is the only place the conversion matters. Convert to kelvin at the start, and convert the answer back at the end if the question expects Celsius.

5. T₁ = 110 K, P₁ = 62 atm; T₂ = 225 K, P₂ = 33 atm, V₂ = 29 L. Find V₁.

V₁ = (P₂V₂T₁) / (P₁T₂) = (33)(29)(110) / [(62)(225)] = 105 270 / 13 950 = 7.5 L

Check: both changes expand the gas — heating it and lowering the pressure — so the starting volume must be much smaller than 29 L. It is, by about a factor of four.

Two habits that make this whole topic safer

1. Kelvin always. The equation is built on absolute temperature; Celsius will silently produce a plausible wrong number.
2. Predict the direction before computing. Every question above has a one-line sanity check, and each one caught a possible inverted ratio. If a change should expand the gas and your answer shrinks it, you have flipped something.

(Small note: question 4 shows "52.0 0C" — that trailing zero is a degree symbol that did not survive copying. It reads as 52.0 °C.)