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. `(1)/(2)RDeltaT`
B. `(1)/(2)KRDeltaT`
C. `(3)/(2)RDeltaT`
D. `(2K)/(3)DeltaT`

Answer

To solve the problem, we need to find the amount of heat absorbed by one mole of an ideal monoatomic gas when its temperature increases by \( \Delta T \) under the condition that \( VT = K \), where \( K \) is a constant.

### Step-by-Step Solution:

1. **Understanding the Process**:
We are given that the volume \( V \) and temperature \( T \) of the gas are related by \( VT = K \). This implies that \( V = \frac{K}{T} \).

2. **Change in Internal Energy**:
For a monoatomic ideal gas, the change in internal energy \( \Delta U \) can be expressed as:
\[ \Delta U = \frac{3}{2} n R \Delta T \]
Since we have 1 mole of gas (\( n = 1 \)):
\[ \Delta U = \frac{3}{2} R \Delta T \]

3. **Work Done by the Gas**:
To find the work done \( W \) during the process, we can use the relation \( P \Delta V \). However, we can also derive it using the ideal gas law and the given relationship:
- From \( VT = K \), we can differentiate to find the relationship between \( dV \) and \( dT \):
\[ V dT + T dV = 0 \implies dV = -\frac{V}{T} dT \]
- The pressure \( P \) can be expressed as:
\[ P = \frac{nRT}{V} = \frac{RT}{K/T} = \frac{RT^2}{K} \]
- The work done by the gas during the temperature change can be calculated as:
\[ W = \int P \, dV \]
However, since we have the relationship \( P V = nRT \), we can also express it in terms of \( T \):
\[ W = \int P \, dV = \int \frac{RT^2}{K} \left(-\frac{V}{T} dT\right) \]
- After evaluating, we find that:
\[ W = -\frac{R}{2} \Delta T \]

4. **Applying the First Law of Thermodynamics**:
The first law of thermodynamics states:
\[ Q = \Delta U + W \]
Substituting the expressions we found:
\[ Q = \frac{3}{2} R \Delta T - \frac{R}{2} \Delta T \]
Simplifying:
\[ Q = \left(\frac{3}{2} R - \frac{1}{2} R\right) \Delta T = R \Delta T \]

5. **Final Result**:
Thus, the amount of heat absorbed by the gas is:
\[ Q = R \Delta T \]