An ideal gas with adiabatic exponent `gamma` is heated at constant pressure. It absorbs Q amount of heat. Fraction of heat absorbed in increasing the temperature is

Options

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

Answer

To solve the problem, we need to find the fraction of heat absorbed by the ideal gas that is used to increase its temperature when heated at constant pressure. Let's break down the solution step by step.

### Step 1: Understand the Heat Absorption at Constant Pressure
When an ideal gas is heated at constant pressure, the heat absorbed (Q) can be expressed using the formula:
\[ Q = n \cdot C_p \cdot \Delta T \]
where:
- \( n \) = number of moles of the gas
- \( C_p \) = molar specific heat at constant pressure
- \( \Delta T \) = change in temperature

### Step 2: Understand the Change in Internal Energy
The change in internal energy (\( \Delta U \)) of the gas can be expressed as:
\[ \Delta U = n \cdot C_v \cdot \Delta T \]
where:
- \( C_v \) = molar specific heat at constant volume

### Step 3: Determine the Fraction of Heat Used to Increase Temperature
The fraction of heat absorbed that goes into increasing the temperature is given by the ratio of the change in internal energy to the total heat absorbed:
\[ \text{Fraction} = \frac{\Delta U}{Q} \]

### Step 4: Substitute the Expressions for \( \Delta U \) and \( Q \)
Substituting the expressions we derived in Steps 1 and 2 into the fraction:
\[ \text{Fraction} = \frac{n \cdot C_v \cdot \Delta T}{n \cdot C_p \cdot \Delta T} \]
Here, \( n \) and \( \Delta T \) cancel out:
\[ \text{Fraction} = \frac{C_v}{C_p} \]

### Step 5: Relate \( C_p \) and \( C_v \) to \( \gamma \)
From thermodynamics, we know that:
\[ \frac{C_p}{C_v} = \gamma \]
Thus, we can express \( \frac{C_v}{C_p} \) as:
\[ \frac{C_v}{C_p} = \frac{1}{\gamma} \]

### Final Result
Therefore, the fraction of heat absorbed in increasing the temperature is:
\[ \text{Fraction} = \frac{1}{\gamma} \]