Two moles of an ideal monoatomic gas undergoes a process `VT =` constant. If temperature of the gas is increased by `DeltaT = 300K`, then the ratio `((DeltaU)/(DeltaQ))` is

Options

A. 2
B. 3
C. 4
D. 6

Explanation

To solve the problem, we need to find the ratio \(\frac{\Delta U}{\Delta Q}\) for a process where \(VT = \text{constant}\) and the temperature of the gas is increased by \(\Delta T = 300K\).

### Step-by-Step Solution:

1. **Identify the type of gas and its properties**:
We have 2 moles of an ideal monoatomic gas. For a monoatomic gas, the degrees of freedom \(f = 3\).

2. **Calculate the change in internal energy (\(\Delta U\))**:
The change in internal energy for an ideal gas is given by the formula:
\[ \Delta U = \frac{f}{2} n R \Delta T \]
For a monoatomic gas, \(f = 3\), and we have \(n = 2\) moles. Therefore:
\[ \Delta U = \frac{3}{2} \times 2 \times R \times 300 \]
Simplifying this, we get:
\[ \Delta U = 3 \times R \times 300 = 900R \]

3. **Determine the work done (\(W\)) during the process**:
For the process \(VT = \text{constant}\), we can use the relation for work done:
\[ W = nR \frac{\Delta T}{1 - \eta} \]
Here, \(\eta = 2\) (since \(VT\) implies \(PV^2 = \text{constant}\)), so:
\[ W = nR \frac{\Delta T}{1 - 2} = nR \frac{300}{-1} = -300nR \]
Substituting \(n = 2\):
\[ W = -600R \]

4. **Calculate the heat added (\(\Delta Q\))**:
According to the first law of thermodynamics:
\[ \Delta Q = \Delta U + W \]
Substituting the values we found:
\[ \Delta Q = 900R - 600R = 300R \]

5. **Calculate the ratio \(\frac{\Delta U}{\Delta Q}\)**:
Now we can find the ratio:
\[ \frac{\Delta U}{\Delta Q} = \frac{900R}{300R} = 3 \]

### Final Answer:
\[ \frac{\Delta U}{\Delta Q} = 3 \]