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

Options

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

Answer

To solve the problem, we will follow the steps outlined below:

### Step 1: Identify Given Values
We have the following values:
- Pressure (P) = 50 N/m²
- Initial Volume (V₁) = 10 m³
- Final Volume (V₂) = 4 m³
- Heat added (Q) = 100 J

### Step 2: Calculate Change in Volume (ΔV)
The change in volume (ΔV) can be calculated as:
\[ \Delta V = V₂ - V₁ = 4 \, \text{m}³ - 10 \, \text{m}³ = -6 \, \text{m}³ \]

### Step 3: Calculate Work Done (W)
The work done on the gas during compression at constant pressure can be calculated using the formula:
\[ W = P \cdot \Delta V \]
Substituting the values:
\[ W = 50 \, \text{N/m}² \cdot (-6 \, \text{m}³) = -300 \, \text{J} \]
(Note: The negative sign indicates that work is done on the gas.)

### Step 4: Apply the First Law of Thermodynamics
According to the first law of thermodynamics:
\[ \Delta Q = \Delta U + W \]
Where:
- ΔQ = Heat added = 100 J
- ΔU = Change in internal energy (unknown)
- W = Work done = -300 J

Rearranging the equation to solve for ΔU:
\[ \Delta U = \Delta Q - W \]
Substituting the known values:
\[ \Delta U = 100 \, \text{J} - (-300 \, \text{J}) = 100 \, \text{J} + 300 \, \text{J} = 400 \, \text{J} \]

### Step 5: Conclusion
The change in internal energy (ΔU) of the gas is:
\[ \Delta U = 400 \, \text{J} \]