Question

Two flasks I and II shown below are connected by a valve of negligible volume. When the valve is opened, the final pressure of the system in bar is \(x \times 10^{-2}\). The value of x is _________. (Integer answer) 
[Assume - Ideal gas; 1 bar = 10\(^5\) Pa; Molar mass of N\(_2\) = 28.0 g mol\(^{-1}\); R=8.31 J mol\(^{-1}\) K\(^{-1}\)] 

 

Hint:

When two gases in non-insulated containers are mixed, a common assumption for finding the final temperature is the conservation of internal energy, leading to \( T_f = (n_1 T_1 + n_2 T_2) / (n_1 + n_2) \). Once the final moles, volume, and temperature are known, the final pressure can be found directly from the Ideal Gas Law.

Answer 1

Correct Answer: 84

Step 1: Understanding the Question:
Two flasks containing nitrogen gas at different conditions are connected. We need to find the final equilibrium pressure after the valve is opened. Since the flasks are not thermally insulated, we assume the system is isolated, so the total internal energy is conserved.
Step 2: Calculate Initial Moles in Each Flask:
Molar mass of N\(_2\) = 28.0 g/mol.
- Flask I: \( n_1 = \frac{2.8 \, \text{g}}{28.0 \, \text{g/mol}} = 0.1 \) mol. T\(_1\) = 300 K, V\(_1\) = 1 L.
- Flask II: \( n_2 = \frac{0.2 \, \text{g}}{28.0 \, \text{g/mol}} \approx 0.00714 \) mol. T\(_2\) = 60 K, V\(_2\) = 2 L.
Step 3: Determine the Final State:
When the valve is opened, the gases mix.
- Total moles, \( n_f = n_1 + n_2 = 0.1 + 0.00714 = 0.10714 \) mol.
- Total volume, \( V_f = V_1 + V_2 = 1 \, \text{L} + 2 \, \text{L} = 3 \, \text{L} = 3 \times 10^{-3} \) m\(^3\).
To find the final temperature, \( T_f \), we use the conservation of internal energy. For an ideal diatomic gas like N\(_2\), \( U = nC_vT = n(\frac{5}{2}R)T \).
\( U_{initial} = U_1 + U_2 = n_1 C_v T_1 + n_2 C_v T_2 \)
\( U_{final} = n_f C_v T_f \)
Setting \( U_{initial} = U_{final} \):
\( n_1 T_1 + n_2 T_2 = n_f T_f \)
\( T_f = \frac{n_1 T_1 + n_2 T_2}{n_f} = \frac{(0.1)(300) + (0.00714)(60)}{0.10714} = \frac{30 + 0.4284}{0.10714} \approx 283.9 \) K.
Step 4: Calculate Final Pressure:
Using the Ideal Gas Law, \( P_f V_f = n_f R T_f \).
\[ P_f = \frac{n_f R T_f}{V_f} \] \[ P_f = \frac{(0.10714 \, \text{mol}) \times (8.31 \, \text{J mol}^{-1} \text{K}^{-1}) \times (283.9 \, \text{K})}{3 \times 10^{-3} \, \text{m}^3} \approx 84260 \, \text{Pa} \] Step 5: Convert to bar and find x:
1 bar = 10\(^5\) Pa.
\[ P_f = \frac{84260}{10^5} = 0.8426 \, \text{bar} \] We are given \( P_f = x \times 10^{-2} \) bar.
\[ 0.8426 = x \times 10^{-2} \] \[ x = 84.26 \] The value of x as an integer is 84.