A gas absorbs 200J heat and undergoes simultaneous expansion against a constant external pressure of `10^(5)` Pa. The volume changes from 4L to 5L. The change in internal energy is

Options

A. 300J
B. 100J
C. `-330J`
D. `-100J`

Explanation

To find the change in internal energy (ΔU) of the gas, we can use the first law of thermodynamics, which states:

\[ \Delta U = Q + W \]

where:
- \( \Delta U \) = change in internal energy
- \( Q \) = heat absorbed by the system
- \( W \) = work done on the system

### Step 1: Identify the values given in the problem.
- Heat absorbed, \( Q = 200 \, \text{J} \)
- External pressure, \( P = 10^5 \, \text{Pa} \)
- Initial volume, \( V_1 = 4 \, \text{L} \)
- Final volume, \( V_2 = 5 \, \text{L} \)

### Step 2: Calculate the work done (W) by the gas during expansion.
The work done by the gas during expansion against a constant external pressure is given by:

\[ W = -P \Delta V \]

where \( \Delta V = V_2 - V_1 \).

Calculating \( \Delta V \):
\[ \Delta V = V_2 - V_1 = 5 \, \text{L} - 4 \, \text{L} = 1 \, \text{L} \]

Now, converting liters to cubic meters:
\[ 1 \, \text{L} = 10^{-3} \, \text{m}^3 \]
Thus,
\[ \Delta V = 1 \, \text{L} = 10^{-3} \, \text{m}^3 \]

Now substituting the values into the work formula:
\[ W = -P \Delta V = - (10^5 \, \text{Pa}) \times (10^{-3} \, \text{m}^3) \]
\[ W = -100 \, \text{J} \]

### Step 3: Substitute the values of Q and W into the first law of thermodynamics equation.
Now we can substitute \( Q \) and \( W \) into the equation for \( \Delta U \):
\[ \Delta U = Q + W \]
\[ \Delta U = 200 \, \text{J} + (-100 \, \text{J}) \]
\[ \Delta U = 200 \, \text{J} - 100 \, \text{J} = 100 \, \text{J} \]

### Final Answer
The change in internal energy \( \Delta U \) is:
\[ \Delta U = 100 \, \text{J} \]

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