In a gaseous system, a gas expands from `10^(-4)m^3` to `2xx10^(-4)m^3` while itspressure remains constant at `10^(-4) Nt//m^2`. Calculate the amount of heat absorbed by the gas inthe expansion. `[gamma=1.67]`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the parameters
Given:
- Initial volume, \( V_i = 10^{-4} \, m^3 \)
- Final volume, \( V_f = 2 \times 10^{-4} \, m^3 \)
- Constant pressure, \( P = 10^{-4} \, N/m^2 \)
- \( \gamma = 1.67 \)

### Step 2: Calculate the work done during expansion
For an isobaric process (constant pressure), the work done \( W \) is given by the formula:
\[ W = P \Delta V \]
where \( \Delta V = V_f - V_i \).

Calculating \( \Delta V \):
\[ \Delta V = V_f - V_i = 2 \times 10^{-4} - 10^{-4} = 10^{-4} \, m^3 \]

Now substituting the values into the work done formula:
\[ W = P \Delta V = 10^{-4} \, N/m^2 \times 10^{-4} \, m^3 = 10^{-8} \, J \]

### Step 3: Calculate the change in internal energy \( \Delta U \)
The change in internal energy for an ideal gas can be calculated using the formula:
\[ \Delta U = n C_V \Delta T \]
where \( C_V \) can be expressed in terms of \( R \) and \( \gamma \):
\[ C_V = \frac{R}{\gamma - 1} \]

Using the ideal gas law \( PV = nRT \), we can find the change in temperature \( \Delta T \):
From the initial and final states:
\[ P V_i = n R T_i \quad \text{(1)} \]
\[ P V_f = n R T_f \quad \text{(2)} \]

Subtracting (1) from (2):
\[ P (V_f - V_i) = n R (T_f - T_i) \]
This gives us:
\[ \Delta T = \frac{P (V_f - V_i)}{n R} \]

Substituting \( W \) for \( P (V_f - V_i) \):
\[ \Delta T = \frac{W}{n R} \]
Substituting \( W = 10^{-8} \, J \):
\[ \Delta T = \frac{10^{-8}}{n R} \]

Now substituting \( \Delta T \) back into the internal energy equation:
\[ \Delta U = n \left(\frac{R}{\gamma - 1}\right) \left(\frac{10^{-8}}{n R}\right) \]
This simplifies to:
\[ \Delta U = \frac{10^{-8}}{\gamma - 1} \]
Substituting \( \gamma = 1.67 \):
\[ \Delta U = \frac{10^{-8}}{1.67 - 1} = \frac{10^{-8}}{0.67} \approx 1.492 \times 10^{-8} \, J \]

### Step 4: Calculate the total heat absorbed \( Q \)
Using the first law of thermodynamics:
\[ Q = \Delta U + W \]
Substituting the values:
\[ Q = 1.492 \times 10^{-8} + 10^{-8} = 2.492 \times 10^{-8} \, J \]

### Final Answer
The amount of heat absorbed by the gas during the expansion is:
\[ Q \approx 2.492 \times 10^{-8} \, J \]

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