A system undergoes a process which absorbed `5 KJ` of heat and undergoing an expansion againt external pressure of `1 atm`, during the process change in internal energy is `300 J`. Then predict the change in volume (lit)

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem step by step, we will 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 on the system

### Given Data:
- Heat absorbed, \(Q = 5 \, \text{kJ} = 5000 \, \text{J}\) (since \(1 \, \text{kJ} = 1000 \, \text{J}\))
- Change in internal energy, \(\Delta U = 300 \, \text{J}\)
- External pressure, \(P = 1 \, \text{atm}\)

### Step 1: Calculate Work Done
From the first law of thermodynamics, we can rearrange the equation to find the work done:

\[ W = \Delta U - Q \]

Substituting the known values:

\[ W = 300 \, \text{J} - 5000 \, \text{J} = -4700 \, \text{J} \]

### Step 2: Relate Work Done to Change in Volume
The work done by the system during expansion against an external pressure is given by:

\[ W = -P \Delta V \]

Where:
- \(P\) is the pressure in atm
- \(\Delta V\) is the change in volume in liters

Since we need to convert the pressure from atm to Joules, we can use the conversion factor:

\[ 1 \, \text{atm} = 101.3 \, \text{J/L} \]

### Step 3: Substitute and Solve for Change in Volume
Now we can substitute \(W\) and \(P\) into the equation:

\[ -4700 \, \text{J} = - (1 \, \text{atm}) \Delta V \]

Substituting the pressure in Joules:

\[ -4700 \, \text{J} = - (101.3 \, \text{J/L}) \Delta V \]

Now, we can solve for \(\Delta V\):

\[ \Delta V = \frac{4700 \, \text{J}}{101.3 \, \text{J/L}} \approx 46.4 \, \text{L} \]

### Final Answer
The change in volume is approximately:

\[ \Delta V \approx 46.4 \, \text{L} \]