5 mol of gas at 5 atmospheric pressure contained in a 100 L cylinder absorbed 30.26 Kj of heat when it expanded to 200 L at 2 atmospheric pressure. The change in the internal energy of gas is

Options

A. `+20.26 J`
B. `+20.26 Kj`
C. `-20.26 KJ`
D. 10 Kj

Explanation

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
- Moles of gas, \(n = 5 \, \text{mol}\)
- Initial pressure, \(P_1 = 5 \, \text{atm}\)
- Initial volume, \(V_1 = 100 \, \text{L}\)
- Final pressure, \(P_2 = 2 \, \text{atm}\)
- Final volume, \(V_2 = 200 \, \text{L}\)
- Heat absorbed, \(Q = 30.26 \, \text{kJ}\)

### Step 2: Calculate the work done (W)
The work done by the gas during expansion can be calculated using the formula:

\[ W = P_2 V_2 - P_1 V_1 \]

First, we need to convert the pressures from atm to L·atm (since 1 L·atm = 101.325 J):

\[ W = (P_2 \cdot V_2) - (P_1 \cdot V_1) \]
\[ W = (2 \, \text{atm} \cdot 200 \, \text{L}) - (5 \, \text{atm} \cdot 100 \, \text{L}) \]
\[ W = 400 \, \text{L·atm} - 500 \, \text{L·atm} \]
\[ W = -100 \, \text{L·atm} \]

Now, convert L·atm to kJ:

\[ W = -100 \, \text{L·atm} \times 0.101325 \, \text{kJ/L·atm} = -10.1325 \, \text{kJ} \approx -10 \, \text{kJ} \]

### Step 3: Calculate the change in internal energy (\(\Delta U\))
Now, substitute \(Q\) and \(W\) into the first law of thermodynamics equation:

\[ \Delta U = Q + W \]
\[ \Delta U = 30.26 \, \text{kJ} - 10 \, \text{kJ} \]
\[ \Delta U = 20.26 \, \text{kJ} \]

### Final Answer
The change in internal energy of the gas is:

\[ \Delta U = 20.26 \, \text{kJ} \]

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