A gas is compressed at a constant pressure of `50N//m^(2)` from a volume `10m^(3)` to a volume of `4m^(3)`. 100J of heat is added to the gas then its internal energy is

Options

A. Increases by 400J
B. Increases by 200J
C. Decreases by 400J
D. Decreases by 200J

Answer

To solve the problem, we will use the first law of thermodynamics, which states:

\[ Q = \Delta U + W \]

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

### Step 1: Identify the given values
From the question, we have:
- Pressure \( P = 50 \, \text{N/m}^2 \)
- Initial Volume \( V_i = 10 \, \text{m}^3 \)
- Final Volume \( V_f = 4 \, \text{m}^3 \)
- Heat added \( Q = 100 \, \text{J} \)

### Step 2: Calculate the work done by the gas
Since the gas is compressed at constant pressure, the work done \( W \) can be calculated using the formula:

\[ W = P \cdot (V_f - V_i) \]

Substituting the values:

\[ W = 50 \, \text{N/m}^2 \cdot (4 \, \text{m}^3 - 10 \, \text{m}^3) \]

Calculating the change in volume:

\[ V_f - V_i = 4 \, \text{m}^3 - 10 \, \text{m}^3 = -6 \, \text{m}^3 \]

Now substituting this back into the work formula:

\[ W = 50 \, \text{N/m}^2 \cdot (-6 \, \text{m}^3) = -300 \, \text{J} \]

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

\[ Q = \Delta U + W \]

Substituting the known values:

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

### Step 4: Solve for the change in internal energy \( \Delta U \)
Rearranging the equation to solve for \( \Delta U \):

\[ \Delta U = 100 \, \text{J} + 300 \, \text{J} = 400 \, \text{J} \]

### Conclusion
The change in internal energy \( \Delta U \) is \( 400 \, \text{J} \).