A gas is compressed at a constant pressure of 50 N/`m^2` from a volume of `10m^3` to a volume of ` 8 m^3`. Energy of 200 J is then added to the gas by heating. Its internal energy is

Options

A. Increases by 300 J
B. Increases by 200 J
C. Increases by 400 J
D. Increases by 100 J

Answer

To find the internal energy of the gas after it has been compressed and heated, we can use the first law of thermodynamics, which states:

\[ \Delta Q = \Delta U + W \]

Where:
- \(\Delta Q\) is the heat added to the system,
- \(\Delta U\) is the change in internal energy,
- \(W\) is the work done on the system.

### Step 1: Identify the given values
- Pressure, \(P = 50 \, \text{N/m}^2\)
- Initial volume, \(V_i = 10 \, \text{m}^3\)
- Final volume, \(V_f = 8 \, \text{m}^3\)
- Heat added, \(\Delta Q = 200 \, \text{J}\)

### Step 2: Calculate the work done on the gas
The work done on the gas during an isobaric process (constant pressure) is given by:

\[ W = P \cdot \Delta V \]

Where \(\Delta V = V_f - V_i\).

Calculating \(\Delta V\):

\[ \Delta V = V_f - V_i = 8 \, \text{m}^3 - 10 \, \text{m}^3 = -2 \, \text{m}^3 \]

Now substituting the values into the work formula:

\[ W = 50 \, \text{N/m}^2 \cdot (-2 \, \text{m}^3) = -100 \, \text{J} \]

### Step 3: Substitute values into the first law of thermodynamics
Now, we can substitute \(\Delta Q\) and \(W\) into the first law of thermodynamics equation:

\[ 200 \, \text{J} = \Delta U + (-100 \, \text{J}) \]

### Step 4: Solve for \(\Delta U\)
Rearranging the equation gives:

\[ \Delta U = 200 \, \text{J} + 100 \, \text{J} = 300 \, \text{J} \]

### Conclusion
The change in internal energy of the gas is:

\[ \Delta U = 300 \, \text{J} \]

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