An ideal gas is initially at temperature T and volume V . Its volume is increased by `DeltaV` due to an increase in temperature `Delta` , pressure remaining constant . The quantity `delta=DeltaV//VDeltaT` varies with temperature as

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Answer

To solve the problem step by step, we will analyze the relationship between volume, temperature, and the given quantities in an isobaric process (constant pressure).

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles of gas
- \( R \) = universal gas constant
- \( T \) = temperature

Since the pressure is constant in this scenario, we can express the relationship between volume and temperature.

### Step 2: Express Volume in Terms of Temperature
From the ideal gas law, if \( P \) is constant, we can write:
\[ V \propto T \]
This means that volume \( V \) is directly proportional to temperature \( T \). We can express this as:
\[ V = kT \]
Where \( k \) is a constant.

### Step 3: Differentiate the Volume Equation
Now, we differentiate both sides of the equation \( V = kT \):
\[ dV = k \, dT \]
This equation shows how a small change in volume relates to a small change in temperature.

### Step 4: Relate Changes in Volume and Temperature
Next, we can write the ratio of changes:
\[ \frac{dV}{V} = \frac{k \, dT}{kT} \]
This simplifies to:
\[ \frac{dV}{V} = \frac{dT}{T} \]

### Step 5: Convert to Finite Changes
Since the problem mentions finite changes (denoted by \( \Delta \)), we can replace \( d \) with \( \Delta \):
\[ \frac{\Delta V}{V} = \frac{\Delta T}{T} \]

### Step 6: Rearranging the Equation
We can rearrange the equation to express the relationship between \( \Delta V \), \( V \), \( \Delta T \), and \( T \):
\[ \Delta V \cdot \Delta T = \frac{V}{T} \]

### Step 7: Define the Quantity \( \delta \)
The problem defines \( \delta \) as:
\[ \delta = \frac{\Delta V}{V \Delta T} \]
Substituting from our previous result:
\[ \delta = \frac{1}{T} \]

### Step 8: Analyze the Relationship
From the equation \( \delta = \frac{1}{T} \), we can see that \( \delta \) is inversely proportional to the temperature \( T \).

### Conclusion
Thus, the quantity \( \delta \) varies inversely with temperature \( T \). If we were to graph this relationship, it would show a hyperbolic decline, indicating that as temperature increases, \( \delta \) decreases.