Calculating Internal Energy Change (ΔE) for a Balloon Expansion

Q1. A balloon is inflated by heating the air inside. The volume changes from $4.00 \times 10^6$ L to $4.50 \times 10^6$ L with the addition of $1.3 \times 10^8$ J of energy as heat. The balloon expands against a constant pressure of 1.0 atm. Calculate $\Delta E$ for the process. (Use $1\text{ L} \cdot \text{atm} = 101.3 \text{ J}$ for unit conversion.)

Background

Topic: Thermodynamics – Internal Energy, Heat, and Work

This question tests your understanding of the First Law of Thermodynamics, specifically how to calculate the change in internal energy ($\Delta E$) when a system absorbs heat and does work against an external pressure.

Key Terms and Formulas

• Internal Energy Change ($\Delta E$): $\Delta E = q + w$

• Heat ($q$): Energy added to the system (positive if absorbed)

• Work ($w$): For expansion against constant pressure: $w = -P \Delta V$

• Pressure ($P$): External pressure (in atm)

• Volume Change ($\Delta V$): $\Delta V = V_{final} - V_{initial}$ (in L)

• Unit Conversion: $1\text{ L} \cdot \text{atm} = 101.3 \text{ J}$

Step-by-Step Guidance

1. Identify the known values: $q = +1.3 \times 10^8$ J, $P = 1.0$ atm, $V_{initial} = 4.00 \times 10^6$ L, $V_{final} = 4.50 \times 10^6$ L.

2. Calculate the change in volume: $\Delta V = V_{final} - V_{initial} = 4.50 \times 10^6 \text{ L} - 4.00 \times 10^6 \text{ L}$.

3. Calculate the work done by the system: $w = -P \Delta V$. Plug in the values for $P$ and $\Delta V$ (in L·atm).

4. Convert the work from L·atm to Joules using $1\text{ L} \cdot \text{atm} = 101.3 \text{ J}$.

5. Set up the equation for $\Delta E$: $\Delta E = q + w$. Substitute the values for $q$ and $w$ (both in Joules).

Try solving on your own before revealing the answer!

Final Answer: $\Delta E = 8 \times 10^7$ J

First, calculate $\Delta V = 0.50 \times 10^6$ L. The work done is $w = -1.0 \text{ atm} \times 0.50 \times 10^6 \text{ L} = -5.0 \times 10^5 \text{ L} \cdot \text{atm}$.

Convert work to Joules: $w = -5.0 \times 10^5 \text{ L} \cdot \text{atm} \times 101.3 \text{ J}/\text{L} \cdot \text{atm} = -5.065 \times 10^7$ J.

Now, $\Delta E = q + w = 1.3 \times 10^8 \text{ J} + (-5.065 \times 10^7 \text{ J}) = 7.93 \times 10^7$ J$, which rounds to $8 \times 10^7$ J.

This result shows the net change in internal energy after accounting for both heat added and work done by the system.