A diatomic ideal gas is expanded at constant pressure. If work done by the system is 10 J then calculated heat absorbed.

Options

A. 25 J
B. 35 J
C. 30 J
D. 40 J

Answer

To solve the problem of calculating the heat absorbed by a diatomic ideal gas that is expanded at constant pressure with a work done of 10 J, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Work done by the system, \( W = 10 \, \text{J} \).
- Since the work is done by the system, we will consider it as negative in our calculations: \( W = -10 \, \text{J} \).

2. **Use the First Law of Thermodynamics:**
The first law of thermodynamics states:
\[ \Delta U = Q + W \]
Where:
- \( \Delta U \) = Change in internal energy
- \( Q \) = Heat absorbed by the system
- \( W \) = Work done by the system

Rearranging the equation to find \( Q \):
\[ Q = \Delta U - W \]

3. **Calculate the Change in Internal Energy (\( \Delta U \)):**
For an ideal gas, the change in internal energy can be expressed as:
\[ \Delta U = n C_V \Delta T \]
However, since we are dealing with a diatomic ideal gas at constant pressure, we can also express \( \Delta U \) in terms of \( C_P \):
\[ \Delta U = n C_P \Delta T \]
For a diatomic ideal gas, the molar heat capacity at constant pressure \( C_P \) is given by:
\[ C_P = \frac{5}{2} R \]

4. **Relate Work Done to Temperature Change:**
Under constant pressure, the work done can also be expressed as:
\[ W = P \Delta V = n R \Delta T \]
From the work done, we can express \( \Delta T \):
\[ -10 = n R \Delta T \implies \Delta T = -\frac{10}{n R} \]

5. **Substituting \( \Delta T \) into \( \Delta U \):**
Now we can substitute \( \Delta T \) into the equation for \( \Delta U \):
\[ \Delta U = n C_P \Delta T = n \left(\frac{5}{2} R\right) \left(-\frac{10}{n R}\right) \]
Simplifying this:
\[ \Delta U = \frac{5}{2} \cdot (-10) = -25 \, \text{J} \]

6. **Calculate Heat Absorbed (\( Q \)):**
Now substituting \( \Delta U \) and \( W \) into the equation for \( Q \):
\[ Q = \Delta U - W = -25 - (-10) = -25 + 10 = -15 \, \text{J} \]

### Final Answer:
The heat absorbed by the system is:
\[ Q = -15 \, \text{J} \]