Temperature of 1 mole of a gas is increased by `2^(@)C` at constant pressure. Work done is :

Options

A. -R
B. -2R
C. `-R//2`
D. -3R

Explanation

To solve the problem of calculating the work done when the temperature of 1 mole of a gas is increased by \(2^\circ C\) at constant pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Work Done Formula**:
The work done (W) on or by the gas at constant pressure can be expressed using the formula:
\[ W = -P \Delta V \]
where \(P\) is the pressure and \(\Delta V\) is the change in volume.

2. **Use the Ideal Gas Law**:
The ideal gas law states:
\[ PV = nRT \]
where \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.

3. **Relate Change in Volume to Change in Temperature**:
At constant pressure, the change in volume can be related to the change in temperature:
\[ P \Delta V = nR \Delta T \]
Therefore, we can express the work done as:
\[ W = -nR \Delta T \]

4. **Substitute Known Values**:
We know:
- \(n = 1\) mole (given)
- \(\Delta T = 2^\circ C = 2 \, \text{K}\) (since the change in temperature in Celsius is equivalent to the change in Kelvin)
- \(R\) is the ideal gas constant.

Substituting these values into the work done formula:
\[ W = -1 \cdot R \cdot 2 \]
\[ W = -2R \]

5. **Conclusion**:
The work done when the temperature of 1 mole of gas is increased by \(2^\circ C\) at constant pressure is:
\[ W = -2R \]

### Final Answer:
The work done is \(-2R\).

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