Show that the volume thermal expansion coefficient for an ideal gas at constant pressure is `(1)/(T)`.

Options

A. `T`
B. `T^(2)`
C. `(1)/(T)`
D. `(1)/(T^(2))`

Answer

To show that the volume thermal expansion coefficient (γ) for an ideal gas at constant pressure is \( \frac{1}{T} \), we can follow these steps:

### Step 1: Write the definition of the volume thermal expansion coefficient
The volume thermal expansion coefficient \( \gamma \) is defined as:
\[ \gamma = \frac{\Delta V}{V \Delta T} \]
where \( \Delta V \) is the change in volume, \( V \) is the original volume, and \( \Delta T \) is the change in temperature.

### Step 2: Express \( \Delta V \) in terms of differential change
We can express the change in volume in terms of a differential:
\[ \gamma = \frac{dV}{V \, dT} \]

### Step 3: Use the ideal gas law
The ideal gas law is given by:
\[ PV = nRT \]
where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature.

### Step 4: Differentiate the ideal gas law with respect to time
Since we are considering constant pressure, we can differentiate the ideal gas equation:
\[ P \, dV = nR \, dT \]
Here, \( P \) is constant, so we can rearrange this to find \( dV \):
\[ dV = \frac{nR}{P} \, dT \]

### Step 5: Substitute \( dV \) into the expression for \( \gamma \)
Now we can substitute \( dV \) into our expression for \( \gamma \):
\[ \gamma = \frac{dV}{V \, dT} = \frac{\frac{nR}{P} \, dT}{V \, dT} \]
The \( dT \) terms cancel out:
\[ \gamma = \frac{nR}{PV} \]

### Step 6: Substitute \( PV \) using the ideal gas law
From the ideal gas law, we know that \( PV = nRT \). We can substitute this into our expression for \( \gamma \):
\[ \gamma = \frac{nR}{nRT} \]

### Step 7: Simplify the expression
The \( nR \) terms cancel out:
\[ \gamma = \frac{1}{T} \]

### Conclusion
Thus, we have shown that the volume thermal expansion coefficient for an ideal gas at constant pressure is:
\[ \gamma = \frac{1}{T} \]

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