A diatomic gas does 80 J of work when expanded isobarically. The heat given to the gas during this process is

Options

A. 700J
B. 350J
C. 280J
D. 300J

Answer

To find the heat given to a diatomic gas during an isobaric expansion where it does 80 J of work, we can follow these steps:

### Step 1: Understand the Process
In an isobaric process, the pressure remains constant while the gas expands. The work done (W) by the gas during this expansion is given as 80 J.

### Step 2: Use the First Law of Thermodynamics
The First Law of Thermodynamics states:
\[ \Delta Q = \Delta U + W \]
where:
- \(\Delta Q\) is the heat added to the system,
- \(\Delta U\) is the change in internal energy,
- \(W\) is the work done by the system.

### Step 3: Calculate the Change in Internal Energy (\(\Delta U\))
For a diatomic ideal gas, the change in internal energy can be calculated using the formula:
\[ \Delta U = n C_v \Delta T \]
For a diatomic gas, \(C_v = \frac{5}{2}R\).

### Step 4: Relate Work Done to Temperature Change
From the work done in an isobaric process:
\[ W = P \Delta V = n R \Delta T \]
We can express \(\Delta T\) in terms of work:
\[ \Delta T = \frac{W}{nR} \]

### Step 5: Substitute \(\Delta T\) into \(\Delta U\)
Now substituting \(\Delta T\) into the equation for \(\Delta U\):
\[ \Delta U = n C_v \Delta T = n \left(\frac{5}{2}R\right) \left(\frac{W}{nR}\right) = \frac{5}{2}W \]
Given \(W = 80 \, J\):
\[ \Delta U = \frac{5}{2} \times 80 = 200 \, J \]

### Step 6: Calculate Heat Added (\(\Delta Q\))
Now substituting \(\Delta U\) and \(W\) back into the First Law equation:
\[ \Delta Q = \Delta U + W = 200 \, J + 80 \, J = 280 \, J \]

### Conclusion
The heat given to the gas during this process is:
\[ \Delta Q = 280 \, J \]

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