If a gas absorbs `200J` of heat and expands by `500 cm^(3)` against a constant pressure of `2 xx 10^(5) N m^(-2)`, then the change in internal energy is

Options

A. `-200 J`
B. `-100 J`
C. `+100 J`
D. `+200J`

Answer

To find the change in internal energy 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 by the system

### Step 1: Identify the given values
From the problem, we have:
- Heat absorbed, \(Q = 200 \, \text{J}\)
- Pressure, \(P = 2 \times 10^5 \, \text{N/m}^2\)
- Volume change, \(\Delta V = 500 \, \text{cm}^3\)

### Step 2: Convert volume change to cubic meters
Since the pressure is given in N/m², we need to convert the volume from cm³ to m³:
\[ \Delta V = 500 \, \text{cm}^3 = 500 \times 10^{-6} \, \text{m}^3 = 5 \times 10^{-4} \, \text{m}^3 \]

### Step 3: Calculate the work done by the gas
The work done \(W\) by the gas during expansion against a constant pressure can be calculated using the formula:
\[ W = P \times \Delta V \]
Substituting the values:
\[ W = (2 \times 10^5 \, \text{N/m}^2) \times (5 \times 10^{-4} \, \text{m}^3) \]
\[ W = 100 \, \text{J} \]

### Step 4: Calculate the change in internal energy
Now, we can substitute the values of \(Q\) and \(W\) into the equation for \(\Delta U\):
\[ \Delta U = Q - W \]
\[ \Delta U = 200 \, \text{J} - 100 \, \text{J} \]
\[ \Delta U = 100 \, \text{J} \]

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

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