Chemistry · General chemistry I · Worked example

Find a gas volume with PV = nRT

What volume does 0.250 mol of nitrogen gas occupy at 25.0 °C and 1.50 atm?

P V = n R T

Convert the temperature to kelvin

25.0 °C + 273.15 = 298.15 K, which is 298.2 K to one decimal place, the precision of 25.0 °C.

T = 25.0 \circ C + 273.15 = 298.2 K

Solve for the unknown

Divide both sides of PV = nRT by P to get V on its own.

V = \frac{n R T}{P}

Substitute with matching units

Use n = 0.250 mol, R = 0.08206 L·atm/(mol·K), T = 298.2 K and P = 1.50 atm. Moles, kelvins and atmospheres cancel, leaving liters.

V = \frac{(0.250 mol) (0.08206 L atm {mol}^{- 1} K^{- 1}) (298.2 K)}{1.50 atm}

\begin{array}{c} V = \frac{(0.250 mol) (298.2 K)}{1.50 atm} \\ \times 0.08206 \frac{L \cdot atm}{mol \cdot K} \end{array}

Calculate and round

The numerator is 6.118, and 6.118 ÷ 1.50 = 4.08 L to three significant figures.

V = 4.08 L

Check against the molar volume

At STP, 0.250 mol would fill 0.250 × 22.414 = 5.60 L. Here the gas is warmer, which expands it, but it is also at 1.50 atm, which compresses it more, so a smaller volume, 4.08 L, makes sense.

Result

The nitrogen occupies 4.08 L.

Your turn

A 5.00 L flask holds 0.200 mol of gas at 27.0 °C. What is the pressure?

Show the answer and explanation

0.985 atm.

T = 27.0 + 273.15 = 300.2 K. P = nRT/V = (0.200)(0.08206)(300.2)/5.00 = 0.985 atm.

P = \frac{(0.200 mol) (0.08206 L atm {mol}^{- 1} K^{- 1}) (300.2 K)}{5.00 L} = 0.985 atm

\begin{array}{c} P = \frac{(0.200 mol) (300.2 K)}{5.00 L} \\ \times 0.08206 \frac{L \cdot atm}{mol \cdot K} \\ = 0.985 atm \end{array}

Keep exploring

Open the gas in Gas laws & mixtures and double the amount to 0.500 mol: the volume doubles, to 8.16 L.

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