A sample of ideal gas `(gamma=1.4)` is heated at constant pressure. If an amount of 100 J heat is supplied to the gas, the work done by the gas is

Options

A. `42.12J`
B. `56.28J`
C. `28.57J`
D. `36.23J`

Answer

To solve the problem of finding the work done by an ideal gas when it is heated at constant pressure, we can follow these steps:

### Step 1: Understand the relationship between heat, work, and internal energy
For an ideal gas, when heat \( Q \) is added at constant pressure, the relationship can be expressed as:
\[ Q = \Delta U + W \]
where:
- \( Q \) is the heat supplied (100 J in this case),
- \( \Delta U \) is the change in internal energy,
- \( W \) is the work done by the gas.

### Step 2: Express the change in internal energy
The change in internal energy for an ideal gas can be expressed as:
\[ \Delta U = n C_v (T_2 - T_1) \]
where \( C_v \) is the molar specific heat at constant volume.

### Step 3: Relate \( C_v \) and \( C_p \)
For an ideal gas, the relationship between \( C_p \) (molar specific heat at constant pressure) and \( C_v \) is given by:
\[ C_p = C_v + R \]
where \( R \) is the gas constant. Additionally, we can relate \( C_v \) to \( C_p \) using the heat capacity ratio \( \gamma \):
\[ \gamma = \frac{C_p}{C_v} \implies C_v = \frac{C_p}{\gamma} \]

### Step 4: Substitute \( C_v \) in the internal energy equation
Substituting \( C_v \) in the equation for \( \Delta U \):
\[ \Delta U = n \left(\frac{C_p}{\gamma}\right) (T_2 - T_1) \]

### Step 5: Substitute \( Q \) into the equation
From the first equation, we can express work done \( W \):
\[ W = Q - \Delta U \]
Substituting the expression for \( \Delta U \):
\[ W = Q - n \left(\frac{C_p}{\gamma}\right) (T_2 - T_1) \]

### Step 6: Use the given values
Given that \( Q = 100 \, J \) and \( \gamma = 1.4 \):
We need to find \( C_p \). For a diatomic ideal gas, \( C_p \) is typically \( \frac{7}{2}R \) and \( C_v \) is \( \frac{5}{2}R \). However, since we are not given the number of moles \( n \) or the specific heat values, we can calculate the work done directly using the relationship:
\[ W = \frac{Q}{\gamma} \left(1 - \frac{1}{\gamma}\right) \]

### Step 7: Calculate the work done
Substituting the values:
\[ W = 100 \left(1 - \frac{1}{1.4}\right) = 100 \left(1 - 0.7143\right) = 100 \times 0.2857 = 28.57 \, J \]

### Conclusion
The work done by the gas is:
\[ W = 28.57 \, J \]