IB Chemistry Ideal gases Practice Questions

Q: What is the molar volume of an ideal gas at STP ($273\,\text{K}$ and $100\,\text{kPa}$)?

$22.7\,\text{dm}^{3}\,\text{mol}^{-1}$. At the IB-defined STP the molar volume of an ideal gas is $22.7\,\text{dm}^{3}\,\text{mol}^{-1}$.

Q: Using $PV = nRT$ with $R = 8.31\,\text{J K}^{-1}\text{mol}^{-1}$, how many moles of gas occupy $2.00\,\text{dm}^{3}$ at $100\,\text{kPa}$ and $300\,\text{K}$?

0.080 mol. Convert: $P = 100000\,\text{Pa}$, $V = 0.00200\,\text{m}^{3}$. $n = \dfrac{PV}{RT} = \dfrac{100000 \times 0.00200}{8.31 \times 300} = \dfrac{200}{2493} = 0.080\,\text{mol}$.

Q: For a fixed mass of gas at constant temperature, the product $P \times _____$ is constant (Boyle's law).

V / volume. Boyle's law: $PV = \text{constant}$ at constant $n$ and $T$.

Q: At STP, $0.25\,\text{mol}$ of an ideal gas occupies _____ $\text{dm}^{3}$ (molar volume $= 22.7\,\text{dm}^{3}\,\text{mol}^{-1}$).

5.68 / 5.675 / 5.7. $V = n \times V_{m} = 0.25 \times 22.7 = 5.675 \approx 5.68\,\text{dm}^{3}$.

Question 1

Which is an assumption of the ideal gas model?

Accepted Answer

Particles have negligible volume and no intermolecular forces. The ideal gas model assumes the particles have negligible volume and that there are no intermolecular forces between them; collisions are perfectly elastic.

Question 2

At constant temperature, how is the pressure of a fixed mass of gas related to its volume?

Accepted Answer

Inversely proportional. Boyle's law states that at constant temperature pressure is inversely proportional to volume: $P \propto \dfrac{1}{V}$.

Question 3

At constant pressure, the kelvin temperature of a fixed mass of gas is doubled. What happens to its volume?

Accepted Answer

It doubles. By Charles's law $V \propto T$ (in kelvin) at constant pressure, so doubling the absolute temperature doubles the volume.

Question 4

What is the molar volume of an ideal gas at STP ($273\,	ext{K}$ and $100\,	ext{kPa}$)?

Accepted Answer

$22.7\,	ext{dm}^{3}\,	ext{mol}^{-1}$. At the IB-defined STP the molar volume of an ideal gas is $22.7\,	ext{dm}^{3}\,	ext{mol}^{-1}$.

Question 5

Using $PV = nRT$ with $R = 8.31\,	ext{J K}^{-1}	ext{mol}^{-1}$, how many moles of gas occupy $2.00\,	ext{dm}^{3}$ at $100\,	ext{kPa}$ and $300\,	ext{K}$?

Accepted Answer

0.080 mol. Convert: $P = 100000\,	ext{Pa}$, $V = 0.00200\,	ext{m}^{3}$. $n = \dfrac{PV}{RT} = \dfrac{100000 	imes 0.00200}{8.31 	imes 300} = \dfrac{200}{2493} = 0.080\,	ext{mol}$.

Question 6

A real gas deviates most from ideal behaviour at:

Accepted Answer

High pressure and low temperature. At high pressure and low temperature the particles are close together and moving slowly, so their finite volume and intermolecular forces become significant and deviation is greatest.

Question 7

For a fixed mass of gas at constant temperature, the product $P 	imes _____$ is constant (Boyle's law).

Accepted Answer

V / volume. Boyle's law: $PV = 	ext{constant}$ at constant $n$ and $T$.

Question 8

At STP, $0.25\,	ext{mol}$ of an ideal gas occupies _____ $	ext{dm}^{3}$ (molar volume $= 22.7\,	ext{dm}^{3}\,	ext{mol}^{-1}$).

Accepted Answer

5.68 / 5.675 / 5.7. $V = n 	imes V_{m} = 0.25 	imes 22.7 = 5.675 \approx 5.68\,	ext{dm}^{3}$.

Question 9

A 6.00 dm3 vessel contains a gas at 7.45 atm and 296 K. Using PV = nRT, the amount of gas is about

Accepted Answer

1.84 mol. n = PV/RT = (7.45 x 6.00)/(0.08206 x 296) = 1.84 mol.

Question 10

For a fixed mass of gas at constant volume, doubling the absolute temperature will

Accepted Answer

double the pressure. At constant V and n, P is proportional to T, so doubling T doubles the pressure.

IB Chemistry: Ideal gases — Practice Questions & Answers

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