Under what conditions will a pure sample of an ideal gas not only exhibit a pressure of 1 atm but al — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

# Finding Conditions for P = 1 atm and c = 1 mol/L **Given Information:** - Pressure: $P = 1$ atm - Concentration: $c = 1$ mol/L - Gas constant: $R = 0.082$ L·atm·mol⁻¹·K⁻¹ **Step 1: Relate concentration to molar volume** Concentration $c = \frac{n}{V}$, so molar volume $V_m = \frac{V}{n} = \frac{1}{c} = \frac{1}{1} = 1$ L/mol **Step 2: Apply ideal gas law** $$PV_m = RT$$ Substituting known values: $$1 \text{ atm} \times 1 \text{ L/mol} = 0.082 \text{ L·atm·mol}^{-1}\text{·K}^{-1} \times T$$ **Step 3: Solve for temperature** $$T = \frac{1}{0.082} = 12.2 \text{ K}$$ **Answer: The conditions are P = 1 atm at T ≈ 12.2 K (or approximately 12 K)** This unusually low temperature makes physical sense: at standard conditions (273 K, 1 atm), ideal gases have much lower concentrations (~0.045 mol/L). To achieve 1 mol/L at just 1 atm requires drastically cooling the gas, which increases the molar density according to the ideal gas equation.