An ideal gas is heated at constant pressure and absorbs amount of heat Q. if the adiabatic exponent is `gamma`. Then find the fraction of heat absorbed in raising the internal energy and performing the work is.

Options

A. `1-1/gamma`
B. `1+1/gamma`
C. `1-2/gamma`
D. `1+2/gamma`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Process
Given that the gas is heated at constant pressure, we know that the heat absorbed \( Q \) will be used to change the internal energy \( \Delta U \) and to do work \( W \).

### Step 2: Write the Expression for Work Done
For a constant pressure process, the work done \( W \) can be expressed as:
\[ W = P \Delta V \]
Using the ideal gas law, we can relate this to temperature change:
\[ W = nR \Delta T \]
where \( n \) is the number of moles, \( R \) is the gas constant, and \( \Delta T \) is the change in temperature.

### Step 3: Write the Expression for Change in Internal Energy
The change in internal energy \( \Delta U \) for an ideal gas is given by:
\[ \Delta U = \frac{F}{2} nR \Delta T \]
where \( F \) is the degrees of freedom of the gas.

### Step 4: Relate \( F \) to the Adiabatic Exponent \( \gamma \)
The adiabatic exponent \( \gamma \) is defined as:
\[ \gamma = \frac{C_p}{C_v} \]
where \( C_p \) and \( C_v \) are the specific heats at constant pressure and volume, respectively. The degrees of freedom \( F \) can be related to \( \gamma \) as follows:
\[ F = 2(\gamma - 1) \]

### Step 5: Substitute \( F \) into the Internal Energy Expression
Substituting \( F \) into the internal energy change:
\[ \Delta U = \frac{2(\gamma - 1)}{2} nR \Delta T = (\gamma - 1) nR \Delta T \]

### Step 6: Calculate the Total Heat Absorbed
The total heat absorbed \( Q \) is the sum of the change in internal energy and the work done:
\[ Q = \Delta U + W = (\gamma - 1) nR \Delta T + nR \Delta T = \gamma nR \Delta T \]

### Step 7: Find the Fraction of Heat Used for Internal Energy Change
The fraction of heat absorbed that goes into raising the internal energy is given by:
\[ \text{Fraction} = \frac{\Delta U}{Q} = \frac{(\gamma - 1) nR \Delta T}{\gamma nR \Delta T} \]
This simplifies to:
\[ \text{Fraction} = \frac{\gamma - 1}{\gamma} \]

### Step 8: Final Result
Thus, the fraction of heat absorbed in raising the internal energy and performing work is:
\[ \text{Fraction} = 1 - \frac{1}{\gamma} \]