Temperature of `1mol` of a gas is increased by `1^(@)` at constant pressure. The work done is

Options

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

Explanation

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

### Step 1: Understand the Work Done Formula
The work done (W) by a gas during expansion or compression at constant pressure is given by the formula:
\[ W = -P \Delta V \]
where \( P \) is the pressure and \( \Delta V \) is the change in volume.

### Step 2: Relate Work Done to Temperature Change
At constant pressure, we can relate the change in volume to the change in temperature using the ideal gas law. The ideal gas law states:
\[ PV = nRT \]
Differentiating this with respect to temperature at constant pressure gives:
\[ P dV = nR dT \]
Thus, we can express the work done as:
\[ W = -P dV = -nR dT \]

### Step 3: Substitute Known Values
In the problem, we have:
- \( n = 1 \) mole (given)
- \( dT = 1 \) °C (which is equivalent to 1 K for calculations in thermodynamics)
- \( R \) is the universal gas constant, approximately \( 8.314 \, \text{J/(mol K)} \)

Now substituting these values into the equation for work done:
\[ W = -nR dT = -1 \times R \times 1 \]
\[ W = -R \]

### Step 4: Final Calculation
Substituting the value of \( R \):
\[ W = -8.314 \, \text{J} \]

### Conclusion
The work done when the temperature of 1 mole of gas is increased by 1°C at constant pressure is:
\[ W = -8.314 \, \text{J} \]