1.00 mole of an ideal gas occupies a volume of 22.4 L at \( 0^{\circ} \mathrm{C} \) and 760 mmHg (STP). It is cooled to \( -41^{\circ} \mathrm{C} \) while the pressure is increased to 805 mmHg . What is the new volume of the gas in liters?

To find the new volume of the gas after it has been cooled and the pressure has been increased, we can use the Ideal Gas Law, which is given by the equation: \[ PV = nRT \] Where: - \( P \) = pressure in atm - \( V \) = volume in liters - \( n \) = number of moles of gas - \( R \) = ideal gas constant (0.0821 L·atm/(K·mol)) - \( T \) = temperature in Kelvin ### Step 1: Convert the initial conditions to Kelvin and atm 1. **Initial Temperature**: \[ T_1 = 0^{\circ}C = 273.15 \, K \] 2. **Initial Pressure**: \[ P_1 = 760 \, mmHg = 1 \, atm \] 3. **Initial Volume**: \[ V_1 = 22.4 \, L \] ### Step 2: Convert the final conditions to Kelvin and atm 1. **Final Temperature**: \[ T_2 = -41^{\circ}C = -41 + 273.15 = 232.15 \, K \] 2. **Final Pressure**: \[ P_2 = 805 \, mmHg = \frac{805}{760} \, atm \approx 1.0592 \, atm \] ### Step 3: Use the Ideal Gas Law to find the new volume Since the number of moles \( n \) and the gas constant \( R \) remain constant, we can use the combined gas law: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] Rearranging to solve for \( V_2 \): \[ V_2 = V_1 \cdot \frac{P_1}{P_2} \cdot \frac{T_2}{T_1} \] ### Step 4: Substitute the known values \[ V_2 = 22.4 \, L \cdot \frac{1 \, atm}{1.0592 \, atm} \cdot \frac{232.15 \, K}{273.15 \, K} \] Calculating each part: 1. Calculate \( \frac{P_1}{P_2} \): \[ \frac{1}{1.0592} \approx 0.9437 \] 2. Calculate \( \frac{T_2}{T_1} \): \[ \frac{232.15}{273.15} \approx 0.8500 \] 3. Now substitute these values back into the equation for \( V_2 \): \[ V_2 = 22.4 \, L \cdot 0.9437 \cdot 0.8500 \] Calculating \( V_2 \): \[ V_2 \approx 22.4 \cdot 0.9437 \cdot 0.8500 \approx 17.7 \, L \] ### Final Answer The new volume of the gas is approximately **17.7 liters**.