A gas is at 1 atm pressure with a volume `800 cm^(3)`. When `100 J` of heat is supplied to the gas, it expands to `1L` at constant pressure. The change in its internal energy is

Options

A. (a)80 J
B. (b)`-80 J`
C. (c)20 J
D. (d)`-20 J`

Answer

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

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

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

### Step 1: Identify the values given in the problem
- Initial pressure \(P = 1 \text{ atm}\)
- Initial volume \(V_i = 800 \text{ cm}^3 = 800 \times 10^{-6} \text{ m}^3\)
- Final volume \(V_f = 1 \text{ L} = 1 \times 10^{-3} \text{ m}^3\)
- Heat supplied \(\Delta Q = 100 \text{ J}\)

### Step 2: Calculate the change in volume \(\Delta V\)
\[ \Delta V = V_f - V_i = (1 \times 10^{-3} \text{ m}^3) - (800 \times 10^{-6} \text{ m}^3) = (1 - 0.8) \times 10^{-3} \text{ m}^3 = 0.2 \times 10^{-3} \text{ m}^3 = 2 \times 10^{-4} \text{ m}^3 \]

### Step 3: Convert pressure from atm to Pascal
\[ 1 \text{ atm} = 1.013 \times 10^5 \text{ Pa} \approx 10^5 \text{ Pa} \]

### Step 4: Calculate the work done \(\Delta W\)
At constant pressure, the work done by the gas during expansion is given by:
\[ \Delta W = P \Delta V \]
Substituting the values:
\[ \Delta W = (10^5 \text{ Pa}) \times (2 \times 10^{-4} \text{ m}^3) = 20 \text{ J} \]

### Step 5: Use the first law of thermodynamics to find \(\Delta U\)
Rearranging the first law:
\[ \Delta U = \Delta Q - \Delta W \]
Substituting the known values:
\[ \Delta U = 100 \text{ J} - 20 \text{ J} = 80 \text{ J} \]

### Final Answer
The change in internal energy \(\Delta U\) is \(80 \text{ J}\).

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