Ideal Gas Law Simulator

The simulator always uses $n = 1.00$ mol and $R = 0.08206\ \text{L atm mol}^{-1}\text{K}^{-1}$. Pick a law from the Law menu, then drag the single slider. A coral dot moves along the curve on the graph and the readout panel updates.

The y-axis tick labels are rounded to whole numbers, so read exact values from the Computed line rather than from the gridlines.

The ideal gas law, with T always in kelvin: $$PV = nRT$$

Holding n and one more variable constant gives each simple law:

• Boyle, constant n and T: $P_1V_1 = P_2V_2$, so $P \propto 1/V$.
• Charles, constant n and P: $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$, so $V \propto T$.
• Gay-Lussac, constant n and V: $\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$, so $P \propto T$.

All three come from the combined form $\dfrac{P_1V_1}{n_1T_1} = \dfrac{P_2V_2}{n_2T_2}$, where you cancel whatever stays the same. Choose R to match your units: $0.08206\ \text{L atm mol}^{-1}\text{K}^{-1}$ with atmospheres and liters, or $8.314\ \text{J mol}^{-1}\text{K}^{-1}$ with pascals and cubic meters. Related forms on the exam include molar mass from density, $M = \dfrac{dRT}{P}$, and Dalton's law of partial pressures, $P_A = P_{total} \cdot X_A$.

Question. 1.00 mol of an ideal gas is held at 300 K in a 10.0 L container. (a) What is its pressure? (b) The gas is compressed to 5.0 L at constant temperature. What is the new pressure? (c) Starting again from 10.0 L and 300 K, the gas is heated to 450 K in the rigid container. What is the pressure?

(a) $P = \dfrac{nRT}{V} = \dfrac{(1.00)(0.08206)(300)}{10.0} = 2.46\ \text{atm}$.

(b) Constant n and T is Boyle's law: $P_2 = \dfrac{P_1V_1}{V_2} = \dfrac{(2.46)(10.0)}{5.0} = 4.92\ \text{atm}$. Halving the volume doubles the pressure.

(c) Constant n and V is Gay-Lussac's law: $P_2 = P_1 \cdot \dfrac{T_2}{T_1} = 2.46 \times \dfrac{450}{300} = 3.69\ \text{atm}$.

Check it in the simulator. Choose Boyle and set V = 10.0 L: the readout shows P = 2.46 atm and the Check line shows PV = 24.62 and nRT = 24.62. Slide to 5.0 L: P = 4.92 atm, and PV is still 24.62. Now choose Gay-Lussac and set T = 450 K: P = 3.69 atm. In Charles mode, 300 K gives V = 24.62 L and 600 K gives 49.24 L, double the volume for double the kelvin temperature.

$PV = nRT$ appears in Unit 3 and again in stoichiometry with gaseous products, in partial-pressure problems, and in equilibrium questions involving $K_p$. Multiple-choice items often ask you to read a graph like the ones here and identify which variables were held constant, or to predict how pressure changes when two variables change at once. The simulator's straight lines and curve are the shapes to recognize.