One mole of diatomic gas is being heated in a closed tank 300K up to 1000K During the process part of the molecules dissociate At 1000K the energy of the diatomic molecules are only half of that of the whole gas By what factor has the gas pressure increased `(P_(final)//P_(initial))` ? (The oscillation of the molecules are not to be taken in account)

Options

A. `160/33`
B. `16/11`
C. `3/2`
D. none

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the dissociation of the gas
We start with 1 mole of diatomic gas (D2) at 300 K. As the temperature increases to 1000 K, some of the diatomic molecules dissociate into monoatomic gas (D). If x moles of diatomic gas dissociate, then we will have:
- Moles of diatomic gas left = \(1 - x\)
- Moles of monoatomic gas produced = \(2x\) (since each mole of diatomic gas produces 2 moles of monoatomic gas)

### Step 2: Set up the energy relationship
At 1000 K, the total energy of the diatomic molecules is half of the total energy of the gas. Therefore, we can express this as:
\[ E_{diatomic} = \frac{1}{2} E_{total} \]
The total energy of the gas is the sum of the energy of the monoatomic and diatomic gases.

### Step 3: Write the energy equations
The energy of the monoatomic gas is given by:
\[ E_{monoatomic} = \frac{3}{2} N_{monoatomic} RT \]
The energy of the diatomic gas is given by:
\[ E_{diatomic} = \frac{5}{2} N_{diatomic} RT \]
Where \(N_{monoatomic} = 2x\) and \(N_{diatomic} = 1 - x\).

### Step 4: Set up the equation based on energy
From the energy relationship, we have:
\[ \frac{3}{2} (2x) R (1000) + \frac{5}{2} (1 - x) R (1000) = 2 \times \frac{5}{2} (1 - x) R (1000) \]
Simplifying this gives:
\[ 3x + 5(1 - x) = 5(1 - x) \]
This leads to:
\[ 3x + 5 - 5x = 5 - 5x \]
\[ 6x = 5 - 5x \]
\[ 11x = 5 \implies x = \frac{5}{11} \]

### Step 5: Calculate the final number of moles
Now we can calculate the final number of moles of gas:
\[ N_{final} = N_{monoatomic} + N_{diatomic} = 2x + (1 - x) = 2 \left(\frac{5}{11}\right) + \left(1 - \frac{5}{11}\right) = \frac{10}{11} + \frac{6}{11} = \frac{16}{11} \]

### Step 6: Calculate initial and final pressures
Using the ideal gas law, we can find the initial and final pressures:
- Initial pressure \(P_i\):
\[ P_i = \frac{N_{initial}RT_{initial}}{V} = \frac{1 \cdot R \cdot 300}{V} = \frac{300R}{V} \]

- Final pressure \(P_f\):
\[ P_f = \frac{N_{final}RT_{final}}{V} = \frac{\frac{16}{11} R \cdot 1000}{V} = \frac{16000R}{11V} \]

### Step 7: Calculate the pressure ratio
Now we can find the ratio of final pressure to initial pressure:
\[ \frac{P_f}{P_i} = \frac{\frac{16000R}{11V}}{\frac{300R}{V}} = \frac{16000}{11 \cdot 300} = \frac{16000}{3300} = \frac{160}{33} \]

### Final Answer
Thus, the factor by which the pressure has increased is:
\[ \frac{P_f}{P_i} = \frac{160}{33} \]