Two moles of an ideal diatomic gas is taken through a process `VT^(2)=` constant so that its temperature increases by `DeltaT=300K`. The ratio `((DeltaU)/(DeltaQ))` of increase in internal energy and heat supplied to the gas during the process is

Options

A. `2`
B. 3
C. 4
D. 5

Explanation

To solve the problem, we need to find the ratio \(\frac{\Delta U}{\Delta Q}\) for the given process where \(VT^2 = \text{constant}\) and the temperature increases by \(\Delta T = 300 \, K\).

### Step 1: Identify the type of process
The process is described by the equation \(VT^2 = \text{constant}\). This indicates that it is a polytropic process. We can express this in terms of pressure and volume as well.

### Step 2: Use the ideal gas law
From the ideal gas law, we have:
\[ PV = nRT \]
For 2 moles of an ideal diatomic gas, we can write:
\[ P = \frac{nRT}{V} \]

### Step 3: Determine the specific heat capacities
For a diatomic gas, the specific heat at constant volume \(C_v\) is given by:
\[ C_v = \frac{5R}{2} \]
The specific heat at constant pressure \(C_p\) is:
\[ C_p = C_v + R = \frac{5R}{2} + R = \frac{7R}{2} \]

### Step 4: Determine the effective specific heat for the process
In a polytropic process, the effective specific heat \(C\) can be calculated using the relation:
\[ C = C_v + \frac{R}{1 - n} \]
where \(n\) is the polytropic index. For the given process, since \(VT^2 = \text{constant}\), we can derive that \(n = \frac{3}{2}\).

Substituting \(n\) into the equation:
\[ C = C_v + \frac{R}{1 - \frac{3}{2}} = C_v - 2R \]
Substituting \(C_v = \frac{5R}{2}\):
\[ C = \frac{5R}{2} - 2R = \frac{5R}{2} - \frac{4R}{2} = \frac{R}{2} \]

### Step 5: Calculate \(\Delta Q\) and \(\Delta U\)
The heat supplied to the gas \(\Delta Q\) can be calculated as:
\[ \Delta Q = nC\Delta T = 2 \cdot \frac{R}{2} \cdot 300 = 300R \]
The change in internal energy \(\Delta U\) can be calculated as:
\[ \Delta U = nC_v\Delta T = 2 \cdot \frac{5R}{2} \cdot 300 = 1500R \]

### Step 6: Calculate the ratio \(\frac{\Delta U}{\Delta Q}\)
Now we can find the ratio:
\[ \frac{\Delta U}{\Delta Q} = \frac{1500R}{300R} = 5 \]

### Final Answer
Thus, the ratio \(\frac{\Delta U}{\Delta Q}\) is \(5\).