Temperature of1 mole of a gas is increased by `1^(@)` at constant pressure. Work done is

Options

A. 2 cal
B. 3 cal
C. 4 cal
D. 5 cal

Explanation

To solve the problem of calculating the work done when the temperature of 1 mole of gas is increased by \(1^\circ C\) 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 the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.

2. **Identify the Initial Conditions**:
Given that we have 1 mole of gas (\(n = 1\)) and the temperature is increased by \(1^\circ C\) (which is equivalent to an increase of \(1 K\) since the size of the degree is the same in Celsius and Kelvin).

3. **Relate the Change in Temperature to Volume**:
When the temperature increases at constant pressure, the volume of the gas will also change. The relationship can be expressed as:
\[ P(V + \Delta V) = nR(T + 1) \]
where \(\Delta V\) is the change in volume.

4. **Substituting Values**:
The initial state can be described as:
\[ PV = nRT \]
Therefore, substituting for \(n = 1\):
\[ PV = RT \]
Now, substituting this into the equation for the new state:
\[ P(V + \Delta V) = RT + R \]

5. **Simplifying the Equation**:
Expanding the left-hand side gives:
\[ PV + P\Delta V = RT + R \]
Since \(PV = RT\), we can substitute:
\[ RT + P\Delta V = RT + R \]
This simplifies to:
\[ P\Delta V = R \]

6. **Work Done Calculation**:
The work done \(W\) by the gas at constant pressure is given by:
\[ W = P\Delta V \]
From our previous step, we found that:
\[ W = R \]

7. **Substituting the Value of R**:
The value of the ideal gas constant \(R\) is approximately \(2 \, \text{calories}\) (or \(8.314 \, \text{J/mol K}\)). Therefore:
\[ W = 2 \, \text{calories} \]

### Final Answer:
The work done when the temperature of 1 mole of gas is increased by \(1^\circ C\) at constant pressure is \(2 \, \text{calories}\).

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