An ideal gas expand against a constant external pressure at 2.0 atmosphere from 20 litre to 40 litre and absorb `10kJ` of energy from surrounding . What is the change in internal energy of the system ?

Options

A. 4052 J
B. 5948 J
C. 14052 J
D. 9940 J

Explanation

To solve the problem, 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 by the system

### Step 1: Identify the Given Values
- External pressure (\(P\)) = 2.0 atm
- Initial volume (\(V_1\)) = 20 L
- Final volume (\(V_2\)) = 40 L
- Heat absorbed (\(Q\)) = 10 kJ = 10,000 J

### Step 2: Calculate the Change in Volume (\(\Delta V\))
\[ \Delta V = V_2 - V_1 = 40 \, \text{L} - 20 \, \text{L} = 20 \, \text{L} \]

### Step 3: Calculate the Work Done (\(W\))
The work done by the system when it expands against a constant external pressure is given by:
\[ W = -P \Delta V \]

First, we need to convert the volume from liters to the appropriate units for work. The pressure is given in atm, and we will convert the volume to liters atmospheric:

\[ W = - (2.0 \, \text{atm}) \times (20 \, \text{L}) = -40 \, \text{L atm} \]

Now, we convert \(W\) from liters-atmospheric to joules using the conversion factor \(1 \, \text{L atm} = 101.325 \, \text{J}\):

\[ W = -40 \, \text{L atm} \times 101.325 \, \text{J/L atm} = -4052 \, \text{J} \]

### Step 4: Substitute Values into the First Law of Thermodynamics
Now we have \(Q\) and \(W\):
- \(Q = 10,000 \, \text{J}\)
- \(W = -4052 \, \text{J}\)

Substituting these values into the first law equation:

\[ \Delta U = Q + W \]
\[ \Delta U = 10,000 \, \text{J} + (-4052 \, \text{J}) = 10,000 \, \text{J} - 4052 \, \text{J} = 5948 \, \text{J} \]

### Final Answer
The change in internal energy of the system is:
\[ \Delta U = 5948 \, \text{J} \]