Heat is supplied to a diatomic gas at constant pressure.
The ratio of `DeltaQ : DeltaU : DeltaW` is

Options

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

Answer

To solve the problem of finding the ratio of \( \Delta Q : \Delta U : \Delta W \) for a diatomic gas at constant pressure, we can follow these steps:

### Step 1: Understand the relationships in an isobaric process
For an isobaric (constant pressure) process, the following relationships hold:
- The heat supplied to the gas, \( \Delta Q \), is given by:
\[ \Delta Q = N C_p \Delta T \]
- The change in internal energy, \( \Delta U \), is given by:
\[ \Delta U = N C_v \Delta T \]
- The work done by the gas, \( \Delta W \), is given by:
\[ \Delta W = P \Delta V = N R \Delta T \]

### Step 2: Determine the specific heat capacities for a diatomic gas
For a diatomic gas, the degrees of freedom \( f \) is 5. The specific heat capacities can be calculated as follows:
- The molar specific heat at constant volume \( C_v \) is:
\[ C_v = \frac{f}{2} R = \frac{5}{2} R \]
- The molar specific heat at constant pressure \( C_p \) is:
\[ C_p = C_v + R = \frac{5}{2} R + R = \frac{7}{2} R \]

### Step 3: Substitute the values into the equations
Now we can substitute \( C_p \) and \( C_v \) into the equations for \( \Delta Q \), \( \Delta U \), and \( \Delta W \):
- For \( \Delta Q \):
\[ \Delta Q = N \left(\frac{7}{2} R\right) \Delta T \]
- For \( \Delta U \):
\[ \Delta U = N \left(\frac{5}{2} R\right) \Delta T \]
- For \( \Delta W \):
\[ \Delta W = N R \Delta T \]

### Step 4: Form the ratio \( \Delta Q : \Delta U : \Delta W \)
Now we can express the ratio \( \Delta Q : \Delta U : \Delta W \):
- From the equations:
\[ \Delta Q : \Delta U : \Delta W = \left(\frac{7}{2} R \Delta T\right) : \left(\frac{5}{2} R \Delta T\right) : \left(R \Delta T\right) \]
- Simplifying this gives:
\[ = 7 : 5 : 2 \]

### Conclusion
Thus, the ratio \( \Delta Q : \Delta U : \Delta W \) for a diatomic gas at constant pressure is:
\[ \Delta Q : \Delta U : \Delta W = 7 : 5 : 2 \]