For an ideal gas , the specific heat at constant pressure `C_p` is greater than the specific heat at constant volume `C_v` This is because

Options

A. There is a finite done by the gas on its environment when its temperature is increased while pressure remains constant
B. There is a finite done by the gas on its environment when its temperature is increased while the volume remains constant
C. The pressure of the gas remains constant when its temperature remains constant
D. The internal energy of the gas at constant pressure is more than at constant volume

Answer

To understand why the specific heat at constant pressure \( C_p \) is greater than the specific heat at constant volume \( C_v \) for an ideal gas, we can analyze the situation step by step.

### Step 1: Understanding Specific Heat
Specific heat is the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). There are two specific heats to consider:
- \( C_v \): Specific heat at constant volume, where the volume of the gas does not change.
- \( C_p \): Specific heat at constant pressure, where the pressure of the gas remains constant.

### Step 2: Applying the First Law of Thermodynamics
The first law of thermodynamics states that:
\[ Q = \Delta U + W \]
where:
- \( Q \) is the heat added to the system,
- \( \Delta U \) is the change in internal energy,
- \( W \) is the work done by the system.

### Step 3: Analyzing Constant Volume Process
In a constant volume process (for \( C_v \)):
- The volume does not change, so no work is done on or by the gas (\( W = 0 \)).
- Therefore, the heat added to the system is equal to the change in internal energy:
\[ Q_v = \Delta U \]
This means that the heat capacity at constant volume is given by:
\[ C_v = \frac{Q_v}{\Delta T} \]

### Step 4: Analyzing Constant Pressure Process
In a constant pressure process (for \( C_p \)):
- The gas can expand, and work is done by the gas as it expands against the external pressure.
- The first law of thermodynamics can be expressed as:
\[ Q_p = \Delta U + W \]
Here, the work done \( W \) is equal to \( P \Delta V \) (where \( P \) is the pressure and \( \Delta V \) is the change in volume).
Thus, we can write:
\[ C_p = \frac{Q_p}{\Delta T} \]

### Step 5: Comparing \( C_p \) and \( C_v \)
From the analysis:
- For constant volume, \( Q_v = \Delta U \).
- For constant pressure, \( Q_p = \Delta U + P \Delta V \).

Since \( P \Delta V \) is always a positive quantity (as the gas does work when it expands), it follows that:
\[ Q_p > Q_v \]
This leads to:
\[ C_p > C_v \]

### Conclusion
Thus, the specific heat at constant pressure \( C_p \) is greater than the specific heat at constant volume \( C_v \) because, at constant pressure, the gas does work on its surroundings, requiring additional heat input to achieve the same temperature change compared to a constant volume process.