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 of finding the ratio \(\frac{\Delta U}{\Delta Q}\) for the given process of an ideal diatomic gas, we can follow these steps:

### Step 1: Identify the Given Information
We have:
- Number of moles, \(n = 2\) moles
- Temperature change, \(\Delta T = 300 \, \text{K}\)
- The process is defined by \(VT^2 = \text{constant}\).

### Step 2: Understand the Process
The equation \(VT^2 = \text{constant}\) suggests that pressure \(P\) is not constant but varies with volume and temperature. This indicates a specific type of process, and we can derive the heat capacities involved.

### Step 3: Determine the Molar Heat Capacity
For a diatomic gas, the molar heat capacity at constant volume \(C_v\) is given by:
\[ C_v = \frac{5R}{2} \]
where \(R\) is the universal gas constant.

### Step 4: Calculate the Molar Heat Capacity for the Process
From the process \(VT^2 = \text{constant}\), we can derive the effective heat capacity \(C\). The relationship can be expressed as:
\[ C = C_v + \frac{R}{1 - X} \]
where \(X\) is the exponent in the volume-temperature relationship. Here, \(X = \frac{3}{2}\) (as derived from the process).

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

### Step 5: Calculate the Heat Supplied (\(\Delta Q\))
The heat supplied to the gas can be calculated using:
\[ \Delta Q = nC\Delta T \]
Substituting the values:
\[ \Delta Q = 2 \times \frac{R}{2} \times 300 = R \times 300 \]

### Step 6: Calculate the Change in Internal Energy (\(\Delta U\))
The change in internal energy for an ideal gas is given by:
\[ \Delta U = nC_v\Delta T \]
Substituting the values:
\[ \Delta U = 2 \times \frac{5R}{2} \times 300 = 5R \times 300 \]

### Step 7: Calculate the Ratio \(\frac{\Delta U}{\Delta Q}\)
Now, we can find the ratio:
\[ \frac{\Delta U}{\Delta Q} = \frac{5R \times 300}{R \times 300} = \frac{5}{1} \]

### Final Answer
The ratio \(\frac{\Delta U}{\Delta Q} = 5\).

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