A gas is heated at a constant pressure. The fraction of heat supplied used of external work is

Options

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

Explanation

To solve the problem, we need to analyze the relationship between the heat supplied to a gas at constant pressure and the work done by the gas. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the First Law of Thermodynamics
The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (ΔQ) minus the work done by the system (ΔW):
\[ \Delta Q = \Delta U + \Delta W \]

### Step 2: Define Heat Supplied at Constant Pressure
For a gas heated at constant pressure, the heat supplied can be expressed as:
\[ \Delta Q = N C_p \Delta T \]
where:
- \( N \) = number of moles of the gas
- \( C_p \) = specific heat capacity at constant pressure
- \( \Delta T \) = change in temperature

### Step 3: Relate Internal Energy Change to Work Done
At constant pressure, the change in internal energy can be expressed as:
\[ \Delta U = N C_v \Delta T \]
where \( C_v \) is the specific heat capacity at constant volume.

### Step 4: Express Work Done
The work done by the gas during expansion at constant pressure is given by:
\[ \Delta W = P \Delta V \]
Using the ideal gas law, we can relate this to temperature changes.

### Step 5: Substitute and Rearrange
From the first law of thermodynamics, we can substitute for ΔQ and ΔU:
\[ N C_p \Delta T = N C_v \Delta T + \Delta W \]
Rearranging gives us:
\[ \Delta W = N C_p \Delta T - N C_v \Delta T \]
\[ \Delta W = N (C_p - C_v) \Delta T \]

### Step 6: Use the Relation Between Cv and Cp
We know that:
\[ C_p - C_v = R \]
where \( R \) is the universal gas constant. Thus:
\[ \Delta W = N R \Delta T \]

### Step 7: Calculate the Fraction of Heat Supplied Used for Work
Now we can find the fraction of heat supplied that is used for external work:
\[ \frac{\Delta W}{\Delta Q} = \frac{N R \Delta T}{N C_p \Delta T} \]
This simplifies to:
\[ \frac{\Delta W}{\Delta Q} = \frac{R}{C_p} \]

### Step 8: Relate \( C_p \) and \( C_v \) to \( \gamma \)
Using the relation \( C_p = \gamma C_v \) and \( C_v = \frac{C_p}{\gamma} \):
\[ \frac{R}{C_p} = 1 - \frac{1}{\gamma} \]

### Final Answer
Thus, the fraction of heat supplied that is used for external work is:
\[ \frac{\Delta W}{\Delta Q} = 1 - \frac{1}{\gamma} \]