A gas is compressed at a constant pressure of 50 `N//m^2` from a volume of 10 `m^2` to a volume of 4`m^2`. Energy of 100 J then added to the gas by heating. Its internal energy is

Options

A. increased by 400 J
B. Increased by 200 J
C. increased by 100 J
D. decreased by 200 J

Answer

To solve the problem, we will use the First Law of Thermodynamics, which states:

\[ \Delta U = Q - W \]

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

### Step 1: Identify the given values
- Pressure, \(P = 50 \, \text{N/m}^2\)
- Initial volume, \(V_1 = 10 \, \text{m}^3\)
- Final volume, \(V_2 = 4 \, \text{m}^3\)
- Heat added, \(Q = 100 \, \text{J}\)

### Step 2: Calculate the change in volume
\[ \Delta V = V_2 - V_1 = 4 \, \text{m}^3 - 10 \, \text{m}^3 = -6 \, \text{m}^3 \]

### Step 3: Calculate the work done by the gas
The work done on the gas during compression at constant pressure is given by:
\[ W = P \cdot \Delta V \]
Substituting the values:
\[ W = 50 \, \text{N/m}^2 \cdot (-6 \, \text{m}^3) = -300 \, \text{J} \]

### Step 4: Apply the First Law of Thermodynamics
Now we can substitute the values into the First Law equation:
\[ \Delta U = Q - W \]
Substituting \(Q\) and \(W\):
\[ \Delta U = 100 \, \text{J} - (-300 \, \text{J}) = 100 \, \text{J} + 300 \, \text{J} = 400 \, \text{J} \]

### Conclusion
The change in internal energy \(\Delta U\) is \(400 \, \text{J}\). Therefore, the internal energy of the gas increases by \(400 \, \text{J}\).

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