To calculate the change in internal energy ($\Delta U$) of the system, we use the first law of thermodynamics:

$\Delta U = q + w$

where:

• $q$ is the heat added to the system (in joules).
• $w$ is the work done on or by the system (in joules).

Step 1: Identify Given Values

• Initial volume: $V_i = 8.15$ L
• Final volume: $V_f = 5.00$ L
• External pressure: $P = 3.80$ atm
• Heat gained by the system: $q = 796$ J

Step 2: Calculate Work Done

Work done by the gas in an isobaric process is given by:

$w = - P \Delta V$

where:
$\Delta V = V_f - V_i = 5.00 - 8.15 = -3.15 \text{ L}$

$w = - (3.80 \text{ atm}) (-3.15 \text{ L})$

$w = 11.97 \text{ L·atm}$

Now, convert L·atm to joules using the conversion factor:
$1 \text{ L·atm} = 101.325 \text{ J}$

$w = 11.97 \times 101.325$

$w = 1213.1 \text{ J}$

Since the volume is decreasing, work is done on the system, making $w$ positive.

Step 3: Calculate Internal Energy Change

$\Delta U = q + w$

$\Delta U = 796 + 1213.1$

$\Delta U = 2009.1 \text{ J}$

Final Answer:

$\Delta U = 2009 \text{ J} \quad \text{(rounded to 3 significant figures)}$

Would you like further clarification or additional details?

Related Questions:

1. What is the first law of thermodynamics, and how does it apply to this problem?
2. How do you determine the sign of work ($w$) in thermodynamic processes?
3. Why do we use the conversion factor $1 \text{ L·atm} = 101.325 \text{ J}$?
4. What happens to internal energy if the system loses heat instead of gaining it?
5. How does this calculation change if the external pressure is not constant?

Tip:

Always check the signs of heat ($q$) and work ($w$). Heat absorbed by the system is positive, and work done on the system is positive, while work done by the system is negative.