Temeprature of `1` mole of a gas is increased by `2^(@)` at constant pressure, then work done is :

Options

A. `R`
B. `2R`
C. `R//2`
D. Data insufficient

Explanation

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

### Step-by-Step Solution:

1. **Understand the Ideal Gas Law**:
The ideal gas law is given by the equation:
\[ PV = nRT \]
where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

2. **Work Done at Constant Pressure**:
The work done \( W \) when a gas expands at constant pressure can be expressed as:
\[ W = P \Delta V \]
where \( \Delta V \) is the change in volume.

3. **Relate Change in Volume to Change in Temperature**:
From the ideal gas law, we can express the change in volume as:
\[ \Delta V = \frac{nR \Delta T}{P} \]
where \( \Delta T \) is the change in temperature.

4. **Substitute Values**:
Given that \( n = 1 \) mole and \( \Delta T = 2 \)°C (which is equivalent to 2 K since the size of the degree is the same), we can substitute these values into the equation:
\[ \Delta V = \frac{1 \cdot R \cdot 2}{P} \]

5. **Calculate Work Done**:
Now substituting \( \Delta V \) back into the work done formula:
\[ W = P \Delta V = P \left(\frac{1 \cdot R \cdot 2}{P}\right) \]
The pressure \( P \) cancels out:
\[ W = 2R \]

6. **Conclusion**:
Therefore, the work done when the temperature of 1 mole of gas is increased by 2°C at constant pressure is:
\[ W = 2R \]

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