`2` moles of a monoatomic gas are expanded to double its initial volume, through a process `(P)//(V) =` constant. If its initial temperature is `300 K`, then which of the following is not true.

Options

A. `triangleT = 900 K`
B. `triangleQ = 3200R`
C. `triangleQ = 3600 R`
D. `W = 900 R`

Explanation

To solve the problem step by step, we will analyze the given information about the monoatomic gas, its expansion, and apply the relevant equations from the kinetic theory of gases.

### Step 1: Understand the Process
We have 2 moles of a monoatomic gas that is expanded to double its initial volume (V2 = 2V1) under the condition that P/V = constant. This implies that the process is a polytropic process where the ratio of pressure to volume remains constant.

### Step 2: Use the Ideal Gas Law
From the ideal gas law, we know:
\[ PV = nRT \]
Where:
- P = pressure
- V = volume
- n = number of moles (2 moles in this case)
- R = universal gas constant
- T = temperature (initially T1 = 300 K)

### Step 3: Relate Pressure and Volume
Since \( P/V = K \) (constant), we can express pressure as:
\[ P = KV \]
Substituting this into the ideal gas equation gives:
\[ KV \cdot V = nRT \]
\[ K V^2 = nRT \]

### Step 4: Find the Relationship Between Temperatures
We can express the relationship between the initial and final states:
\[ K V_1^2 = nRT_1 \]
\[ K V_2^2 = nRT_2 \]
Since \( V_2 = 2V_1 \):
\[ K (2V_1)^2 = nRT_2 \]
\[ 4KV_1^2 = nRT_2 \]
Dividing the two equations:
\[ \frac{4KV_1^2}{KV_1^2} = \frac{nRT_2}{nRT_1} \]
This simplifies to:
\[ 4 = \frac{T_2}{T_1} \]
Thus:
\[ T_2 = 4T_1 \]

### Step 5: Calculate Final Temperature
Given \( T_1 = 300 K \):
\[ T_2 = 4 \times 300 = 1200 K \]

### Step 6: Calculate Change in Temperature
The change in temperature (ΔT) is:
\[ \Delta T = T_2 - T_1 = 1200 K - 300 K = 900 K \]

### Step 7: Calculate Work Done
The work done (W) during the expansion can be calculated using:
\[ W = \int P dV \]
Since \( P = KV \), we can substitute this into the integral:
\[ W = \int KV dV = K \int V dV = K \left[ \frac{V^2}{2} \right]_{V_1}^{V_2} \]
Calculating this gives:
\[ W = K \left( \frac{(2V_1)^2}{2} - \frac{V_1^2}{2} \right) = K \left( \frac{4V_1^2}{2} - \frac{V_1^2}{2} \right) = K \left( \frac{3V_1^2}{2} \right) \]

Using \( K V_1^2 = nRT_1 \), we can express the work done as:
\[ W = \frac{3}{2} nRT_1 \]

Substituting \( n = 2 \) and \( T_1 = 300 K \):
\[ W = \frac{3}{2} \times 2R \times 300 = 900R \]

### Step 8: Calculate Change in Internal Energy
The change in internal energy (ΔU) for a monoatomic gas is given by:
\[ \Delta U = nC_V \Delta T \]
Where \( C_V = \frac{3R}{2} \) for a monoatomic gas:
\[ \Delta U = n \left( \frac{3R}{2} \right) \Delta T \]
Substituting \( n = 2 \) and \( \Delta T = 900 K \):
\[ \Delta U = 2 \left( \frac{3R}{2} \right) \times 900 = 2700R \]

### Step 9: Calculate Heat Transfer (ΔQ)
Using the first law of thermodynamics:
\[ \Delta Q = \Delta U + W \]
Substituting the values:
\[ \Delta Q = 2700R + 900R = 3600R \]

### Conclusion
Now we can summarize the results:
1. Change in temperature \( \Delta T = 900 K \)
2. Work done \( W = 900R \)
3. Change in internal energy \( \Delta U = 2700R \)
4. Heat transfer \( \Delta Q = 3600R \)

The question asks which statement is not true. Based on our calculations, if any statement contradicts these results, it is the one that is not true.