<u>Answer:</u> The change in internal energy of the gas is 29.414 kJ.
<u>Explanation:</u>
To calculate the temperature of the gas at different volumes, we use ideal gas equation:

- When volume =

We are given:
Conversion used: 

Putting values in above equation:

- When volume =

We are given:

Putting values in above equation:

- To calculate the change in internal energy, we use the equation:

where,
= change in internal energy = ?
n = number of moles = 4.0 mol
= heat capacity at constant volume = 
= final temperature = 1095.06 K
= initial temperature = 505.41 K
Putting values in above equation, we get:

Converting this value in kilojoules, we use the conversion factor:
1 kJ = 1000 J
So, 
Hence, the change in internal energy of the gas is 29.414 kJ.