Problem 115 A flask containing 2.0 moles of He gas at \( 1.0 \mathrm{~atm} \) and \( 300 \mathrm{~K} \) is connected to another flask containing \( \mathrm{N}_{2}(g) \) at the same temperature and pressure by a narrow tube of negligible volume. Volume of the nitrogen flask is three times volume of He-flask. Now the He-flask is placed in a thermostat at \( 200 \mathrm{~K}_{\text {and }} \mathrm{N}_{2} \)-flask in another thermostat at \( 400 \mathrm{~K} \). Determine final pressure and final number of moles in each flask.

Step 1 ：1. Use ideal gas law to find volume of He flask: \( V_{He} = \frac{n_{He}RT}{P} \)

Step 2 ：2. Calculate volume of N2 flask: \( V_{N2} = 3 * V_{He} \)

Step 3 ：3. Determine total volume after temperature change: \( V_{He}^{'} = \frac{n_{He}R(200)}{1.0} \)

Step 4 ：\( V_{N2}^{'} = \frac{3n_{He}R(400)}{1.0} \)

Step 5 ：4. Calculate final pressure: \( P_{final} = \frac{n_{He}RT_{1}+3n_{He}RT_{2}}{V_{He}^{'}+V_{N2}^{'}} \)

Step 6 ：5. Update the number of moles: \( n_{He}^{'} = \frac{2}{1+n} \)

Step 7 ：\( n_{N2}^{'} = \frac{2n}{1+n} \)

Step 8 ：6. Solve for n: \( n = \frac{3(400)-200}{300} - 1 \)