In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT= K, where K is a constant. In this process the temperataure of the gas is increased by `DeltaT`. The amount of heat absorbed by gas is (R is gas constant).

Options

A. `(2K)/3 Detla T`
B. `1/2 R Delta T`
C. `3/2 R Delta T`
D. `1/2 KR Delta T`

Explanation

To solve the problem, we need to analyze the given relationship between temperature (T) and volume (V) of one mole of an ideal monoatomic gas, which is described by the equation \( VT = K \), where \( K \) is a constant. We are tasked with finding the amount of heat absorbed by the gas when the temperature increases by \( \Delta T \).

### Step-by-Step Solution:

1. **Understand the relationship**:
From the relation \( VT = K \), we can express volume \( V \) in terms of temperature \( T \):
\[ V = \frac{K}{T} \]

2. **Identify the type of process**:
The relation \( VT = K \) indicates that this is a polytropic process. For a polytropic process, we can use the formula for the heat capacity \( C \) given by:
\[ C = C_v + \frac{R}{1 - n} \]
where \( n \) is the polytropic index.

3. **Determine the polytropic index**:
From the relation \( VT = K \), we can derive that:
\[ PV = nRT \implies P = \frac{nRT}{V} \]
Substituting \( V = \frac{K}{T} \):
\[ P = \frac{nRT^2}{K} \]
Since \( V \) and \( T \) are inversely related, we can identify that the process is similar to an isothermal process, but we need to find the specific heat for our case.

4. **Calculate the heat absorbed**:
The heat absorbed \( \Delta Q \) during the process can be expressed as:
\[ \Delta Q = C \Delta T \]
We need to find \( C \) for our specific process. For a monoatomic ideal gas, the molar heat capacity at constant volume \( C_v \) is:
\[ C_v = \frac{3R}{2} \]
For our polytropic process, we can assume \( n = 1 \) (since \( V \propto \frac{1}{T} \)), leading to:
\[ C = C_v + R = \frac{3R}{2} + R = \frac{5R}{2} \]

5. **Substituting into the heat equation**:
Now substituting \( C \) back into the heat equation:
\[ \Delta Q = C \Delta T = \frac{5R}{2} \Delta T \]

6. **Final Result**:
Therefore, the amount of heat absorbed by the gas when the temperature increases by \( \Delta T \) is:
\[ \Delta Q = \frac{5R}{2} \Delta T \]