At approximately what temperature will 40. g of argon gas at 2.0 atm occupy a volume of 22.4 L?

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The molar mass of argon gas is approximately 40 g. Therefore, we have 1 mole of argon gas at 2.0 atm and 22.4 L.

We know that for STP, 1 mole of argon gas occupies 22.4 L at 1 atm and 273 K.

Because temperature and pressure are directly proportional, doubling the pressure will double the temperature.

$$P \propto T$$ $$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$ $$\frac{1 \text{ atm}}{273 \text{ K}} = \frac{2 \text{ atm}}{T_2}$$ $$T_2 = 2(273 \text{ K})$$ $$T_2 \approx \boxed{550 \text{ K}}$$

Although harder for the non-calculator section, we can use the ideal gas law.

$$PV=nRT$$ $$(2.0 \text{ atm})(22.4 L) = (1 \text{ mol})\left(0.0821 \frac{\text{ atm}\cdot L}{\text{ mol}\cdot \text{ K}}\right)(T)$$ $$T = \frac{44}{0.0821}$$ $$T \approx \boxed{550 \text{ K}}$$