Chemistry · General chemistry I · Worked example

Use the combined gas law

A weather balloon holds 2.50 L of helium at 1.00 atm and 25.0 °C. What is its volume at 0.500 atm and −15.0 °C?

\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}

List both states, in kelvin

Convert both temperatures: 25.0 °C is 298.2 K and −15.0 °C is 258.2 K. The final volume is the unknown.

The two states of the helium
State	P (atm)	V (L)	T (K)
Initial	1.00	2.50	298.2
Final	0.500	?	258.2

Solve for V₂

The amount of helium is fixed, so P₁V₁/T₁ = P₂V₂/T₂. Multiply both sides by T₂ and divide by P₂.

V_{2} = \frac{P_{1} V_{1} T_{2}}{P_{2} T_{1}}

Substitute and calculate

Halving the pressure doubles the volume, and cooling shrinks it by the factor 258.2/298.2.

V_{2} = \frac{(1.00) (2.50) (258.2)}{(0.500) (298.2)} L = 4.33 L

\begin{array}{c} V_{2} = \frac{(1.00) (2.50) (258.2)}{(0.500) (298.2)} L \\ = 4.33 L \end{array}

Check the direction

Lower pressure expands the gas and lower temperature contracts it. The pressure change is the larger effect, so the volume grows, from 2.50 L to 4.33 L.

Result

The balloon’s volume becomes 4.33 L.

Your turn

A gas occupies 3.00 L at 20.0 °C. At constant pressure, what is its volume at 80.0 °C?

Show the answer and explanation

3.61 L.

At constant pressure, V₂ = V₁T₂/T₁ = 3.00 L × 353.2 K/293.2 K = 3.61 L, which is Charles’s law. Using Celsius would wrongly give 12.0 L.

V_{2} = 3.00 L \times \frac{353.2 K}{293.2 K} = 3.61 L

\begin{array}{c} V_{2} = 3.00 L \times \frac{353.2 K}{293.2 K} \\ = 3.61 L \end{array}

Keep exploring

Open it in Gas laws & mixtures and set the final temperature to 25.0 °C: the volume exactly doubles, to 5.00 L. That is Boyle’s law.

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