The ratio of work done by an ideal diatomic gas to the heat supplied by the gas in an isobaric process is

Options

A. `5/7`
B. `3/5`
C. `2/7`
D. `5/3`

Explanation

To find the ratio of work done by an ideal diatomic gas to the heat supplied by the gas in an isobaric process, we can follow these steps:

### Step 1: Understand the First Law of Thermodynamics
The first law of thermodynamics states:
\[ \Delta Q = \Delta U + \Delta W \]
Where:
- \(\Delta Q\) is the heat supplied to the system,
- \(\Delta U\) is the change in internal energy,
- \(\Delta W\) is the work done by the system.

### Step 2: Determine the Change in Internal Energy (\(\Delta U\))
For an ideal diatomic gas, the change in internal energy can be expressed as:
\[ \Delta U = n C_v \Delta T \]
Where:
- \(n\) is the number of moles,
- \(C_v\) is the molar specific heat at constant volume,
- \(\Delta T\) is the change in temperature.

For a diatomic gas, the degrees of freedom \(f\) is 5, so:
\[ C_v = \frac{f}{2} R = \frac{5}{2} R \]
Thus, we can write:
\[ \Delta U = n \left(\frac{5}{2} R\right) \Delta T = \frac{5}{2} n R \Delta T \]

### Step 3: Determine the Heat Supplied (\(\Delta Q\))
For an isobaric process, the heat supplied is given by:
\[ \Delta Q = n C_p \Delta T \]
Where:
- \(C_p\) is the molar specific heat at constant pressure.

For a diatomic gas:
\[ C_p = C_v + R = \frac{5}{2} R + R = \frac{7}{2} R \]
Thus, we can write:
\[ \Delta Q = n \left(\frac{7}{2} R\right) \Delta T = \frac{7}{2} n R \Delta T \]

### Step 4: Determine the Work Done (\(\Delta W\))
From the first law of thermodynamics, we can express the work done as:
\[ \Delta W = \Delta Q - \Delta U \]
Substituting the expressions for \(\Delta Q\) and \(\Delta U\):
\[ \Delta W = \frac{7}{2} n R \Delta T - \frac{5}{2} n R \Delta T \]
\[ \Delta W = \left(\frac{7}{2} - \frac{5}{2}\right) n R \Delta T = \frac{2}{2} n R \Delta T = n R \Delta T \]

### Step 5: Calculate the Ratio \(\frac{\Delta W}{\Delta Q}\)
Now we can find the ratio of work done to heat supplied:
\[ \frac{\Delta W}{\Delta Q} = \frac{n R \Delta T}{\frac{7}{2} n R \Delta T} \]
The \(n R \Delta T\) terms cancel out:
\[ \frac{\Delta W}{\Delta Q} = \frac{1}{\frac{7}{2}} = \frac{2}{7} \]

### Final Answer
The ratio of work done by an ideal diatomic gas to the heat supplied by the gas in an isobaric process is:
\[ \frac{2}{7} \]