A rigid and insulated tank of `3m^(3)` volume is divided into two compartments. One compartment of volume of `2m^(3)` contains an ideal gas at `0.8314` Mpa and 400 K while the second compartment of volume of `1m^(3)` contains the same gas at `8.314` Mpa and 500 K. If the partition between the two compartments is rptured, the final temperature of the gas is :

Options

A. 420 K
B. 450 K
C. 480 K
D. None of these

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Write down the given data
- For compartment 1:
- Volume (V1) = 2 m³
- Pressure (P1) = 0.8314 MPa = 0.8314 × 10^6 Pa
- Temperature (T1) = 400 K

- For compartment 2:
- Volume (V2) = 1 m³
- Pressure (P2) = 8.314 MPa = 8.314 × 10^6 Pa
- Temperature (T2) = 500 K

### Step 2: Calculate the number of moles in each compartment using the ideal gas equation
The ideal gas equation is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = gas constant = 8.314 J/(mol·K)
- \( T \) = temperature

#### For compartment 1:
\[ n_1 = \frac{P_1 V_1}{RT_1} \]
Substituting the values:
\[ n_1 = \frac{(0.8314 \times 10^6 \, \text{Pa}) \times (2 \, \text{m}^3)}{(8.314 \, \text{J/(mol·K)}) \times (400 \, \text{K})} \]

Calculating:
\[ n_1 = \frac{1662680}{3325.6} \approx 500 \, \text{moles} \]

#### For compartment 2:
\[ n_2 = \frac{P_2 V_2}{RT_2} \]
Substituting the values:
\[ n_2 = \frac{(8.314 \times 10^6 \, \text{Pa}) \times (1 \, \text{m}^3)}{(8.314 \, \text{J/(mol·K)}) \times (500 \, \text{K})} \]

Calculating:
\[ n_2 = \frac{8314000}{4157} \approx 2000 \, \text{moles} \]

### Step 3: Apply the first law of thermodynamics
Since the tank is insulated and rigid, the heat transfer \( Q = 0 \) and work done \( W = 0 \). Therefore, the change in internal energy \( \Delta U = 0 \).

The change in internal energy can be expressed as:
\[ \Delta U = n_1 C_v (T_f - T_1) + n_2 C_v (T_f - T_2) \]
Since \( C_v \) is the same for both compartments, we can factor it out:
\[ 0 = C_v \left( n_1 (T_f - T_1) + n_2 (T_f - T_2) \right) \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ n_1 (T_f - T_1) + n_2 (T_f - T_2) = 0 \]

Expanding this:
\[ n_1 T_f - n_1 T_1 + n_2 T_f - n_2 T_2 = 0 \]
\[ (n_1 + n_2) T_f = n_1 T_1 + n_2 T_2 \]

### Step 5: Solve for the final temperature \( T_f \)
Substituting the values:
\[ T_f = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2} \]
\[ T_f = \frac{(500 \times 400) + (2000 \times 500)}{500 + 2000} \]
\[ T_f = \frac{200000 + 1000000}{2500} \]
\[ T_f = \frac{1200000}{2500} \]
\[ T_f = 480 \, \text{K} \]

### Final Answer:
The final temperature of the gas after the partition is ruptured is **480 K**.

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