Find the ratio of `(DeltaQ)/(DeltaU)` and `(DeltaQ)/(DeltaW)` in an isobaric process. The ratio of molar heat capacities `(C_p)/(C_V)=gamma`.

Explanation

To solve the problem, we need to find the ratios \(\frac{\Delta Q}{\Delta U}\) and \(\frac{\Delta Q}{\Delta W}\) for an isobaric process, given that the ratio of molar heat capacities \(\frac{C_p}{C_V} = \gamma\).

### Step-by-Step Solution:

1. **Understanding the Isobaric Process:**
In an isobaric process, the pressure remains constant. The heat added to the system (\(\Delta Q\)) can be expressed in terms of the molar heat capacity at constant pressure (\(C_p\)) and the change in temperature (\(\Delta T\)):
\[ \Delta Q = n C_p \Delta T \]

2. **Finding \(\Delta U\):**
The change in internal energy (\(\Delta U\)) for an ideal gas can be expressed in terms of the molar heat capacity at constant volume (\(C_V\)):
\[ \Delta U = n C_V \Delta T \]

3. **Calculating the Ratio \(\frac{\Delta Q}{\Delta U}\):**
Now, we can find the ratio of heat added to the change in internal energy:
\[ \frac{\Delta Q}{\Delta U} = \frac{n C_p \Delta T}{n C_V \Delta T} \]
The \(n\) and \(\Delta T\) terms cancel out, so we have:
\[ \frac{\Delta Q}{\Delta U} = \frac{C_p}{C_V} \]
Given that \(\frac{C_p}{C_V} = \gamma\), we find:
\[ \frac{\Delta Q}{\Delta U} = \gamma \]

4. **Finding \(\Delta W\):**
The work done (\(\Delta W\)) in an isobaric process can be expressed as:
\[ \Delta W = P \Delta V = n R \Delta T \]
where \(R\) is the universal gas constant.

5. **Calculating the Ratio \(\frac{\Delta Q}{\Delta W}\):**
Now, we can find the ratio of heat added to the work done:
\[ \frac{\Delta Q}{\Delta W} = \frac{n C_p \Delta T}{n R \Delta T} \]
Again, the \(n\) and \(\Delta T\) terms cancel out:
\[ \frac{\Delta Q}{\Delta W} = \frac{C_p}{R} \]

6. **Using the Relation Between Heat Capacities:**
We know that:
\[ C_p = C_V + R \]
Therefore:
\[ \frac{C_p}{R} = \frac{C_V + R}{R} = \frac{C_V}{R} + 1 \]

7. **Expressing \(\frac{\Delta Q}{\Delta W}\) in terms of \(\gamma\):**
Using the relation \(\frac{C_p}{C_V} = \gamma\), we can express \(C_V\) in terms of \(\gamma\) and \(R\):
\[ C_V = \frac{R}{\gamma - 1} \]
Substituting this into our equation gives:
\[ \frac{\Delta Q}{\Delta W} = \frac{C_V}{R} + 1 = \frac{1}{\gamma - 1} + 1 = \frac{\gamma}{\gamma - 1} \]

### Final Results:
- \(\frac{\Delta Q}{\Delta U} = \gamma\)
- \(\frac{\Delta Q}{\Delta W} = \frac{\gamma}{\gamma - 1}\)