Two vessels `A` and `B`, thermally insulated, contain an ideal monoatomic gas. A small tube fitted with a valve connects these vessels. Initially the vessel `A` has `2` litres of gas at `300K` and `2 xx 10^(5)Nm^(-2)` pressure while vessel `B` has `4` litres of gas at `350 K` and `4 xx 10^(5)Nm^(-2)` pressure. The value is now opened and the system reaches equilibrium in pressure and temperature. Calculate the new pressure and temperature . `(R = (25)/(3)J//mol-K)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- For vessel A:
- Volume \( V_A = 2 \, \text{litres} = 2 \times 10^{-3} \, \text{m}^3 \)
- Pressure \( P_A = 2 \times 10^5 \, \text{N/m}^2 \)
- Temperature \( T_A = 300 \, \text{K} \)

- For vessel B:
- Volume \( V_B = 4 \, \text{litres} = 4 \times 10^{-3} \, \text{m}^3 \)
- Pressure \( P_B = 4 \times 10^5 \, \text{N/m}^2 \)
- Temperature \( T_B = 350 \, \text{K} \)

- Universal gas constant \( R = \frac{25}{3} \, \text{J/(mol K)} \)

### Step 2: Calculate the number of moles in each vessel using the ideal gas equation
The ideal gas equation is given by:
\[ PV = nRT \]
Rearranging gives:
\[ n = \frac{PV}{RT} \]

#### For vessel A:
\[ N_A = \frac{P_A V_A}{RT_A} = \frac{(2 \times 10^5) \times (2 \times 10^{-3})}{\left(\frac{25}{3}\right) \times 300} \]
Calculating \( N_A \):
\[ N_A = \frac{4 \times 10^2}{\frac{25}{3} \times 300} = \frac{400}{2500} = \frac{4}{25} \, \text{mol} \]

#### For vessel B:
\[ N_B = \frac{P_B V_B}{RT_B} = \frac{(4 \times 10^5) \times (4 \times 10^{-3})}{\left(\frac{25}{3}\right) \times 350} \]
Calculating \( N_B \):
\[ N_B = \frac{16 \times 10^2}{\frac{25}{3} \times 350} = \frac{1600}{8750} = \frac{96}{175} \, \text{mol} \]

### Step 3: Calculate the total number of moles and total volume after mixing
The total number of moles \( N \) is:
\[ N = N_A + N_B = \frac{4}{25} + \frac{96}{175} \]
Finding a common denominator (175):
\[ N = \frac{28}{175} + \frac{96}{175} = \frac{124}{175} \, \text{mol} \]

The total volume after mixing:
\[ V_{\text{total}} = V_A + V_B = 2 \, \text{litres} + 4 \, \text{litres} = 6 \, \text{litres} = 6 \times 10^{-3} \, \text{m}^3 \]

### Step 4: Calculate the temperature of the mixture
Using the formula for internal energy:
\[ U = N_A C_V T_A + N_B C_V T_B \]
Where \( C_V = \frac{3R}{2} \).

The temperature of the mixture \( T \) can be calculated as:
\[ N C_V T = N_A C_V T_A + N_B C_V T_B \]
Cancelling \( C_V \):
\[ T = \frac{N_A T_A + N_B T_B}{N} \]
Substituting the values:
\[ T = \frac{\left(\frac{4}{25} \times 300\right) + \left(\frac{96}{175} \times 350\right)}{\frac{124}{175}} \]
Calculating:
\[ T = \frac{\left(\frac{1200}{25}\right) + \left(\frac{3360}{175}\right)}{\frac{124}{175}} = \frac{48 + 19.2}{0.711} \approx 338.7 \, \text{K} \]

### Step 5: Calculate the pressure of the mixture using the ideal gas equation
Using the ideal gas equation:
\[ P = \frac{nRT}{V} \]
Substituting the values:
\[ P = \frac{\left(\frac{124}{175}\right) \left(\frac{25}{3}\right) (338.7)}{6 \times 10^{-3}} \]
Calculating:
\[ P = \frac{124 \times 25 \times 338.7}{175 \times 3 \times 6 \times 10^{-3}} = \frac{1042500}{3150} \approx 330.95 \times 10^3 \, \text{N/m}^2 \approx \frac{10}{3} \times 10^5 \, \text{N/m}^2 \]

### Final Answer
- New Pressure \( P \approx \frac{10}{3} \times 10^5 \, \text{N/m}^2 \)
- New Temperature \( T \approx 338.7 \, \text{K} \)