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Understanding the First Law of Thermodynamics is key to grasping the changes occurring in our helium-filled balloon. This law is fundamentally about the conservation of energy. It states that the change in internal energy of a system, denoted as \( \Delta E \), is equal to the heat added to the system \( q \) plus the work done on the system \( w \). This can be summarized with the equation:
\[ \Delta E = q + w \]
In simpler terms, any energy gained or lost by the system must be accounted for by heat transfer or work done on or by the system. In our balloon example, we calculated \( q \) as the heat absorbed due to a rise in temperature. Since the balloon expands, work is done by the system as it pushes against the external pressure. The calculated work \( w \) is negative, signifying energy leaving the system as the balloon expands. At the end of the problem, combining \( q \) and \( w \) gives us \( \Delta E \), the net change in internal energy of the helium gas.

Molar heat capacity is an essential concept in understanding how substances store and transfer heat. It represents the amount of heat needed to raise the temperature of one mole of a substance by one degree Celsius. For helium gas, the molar heat capacity \( C_p \) given is \( 20.8 \text{ J/掳C路mol} \).
This value is used to calculate the heat absorbed by the system when the temperature changes. In this exercise, the temperature of the helium increases from \(0.0^{\circ} \text{C} \) to \(38.0^{\circ} \text{C} \). Thus, the heat \( q \) can be calculated using the formula:
\[ q = n C_p \Delta T \]
Here, \( n \) is the number of moles, and \( \Delta T \) is the change in temperature. Applying these values gives the amount of heat input into the system, which is \( q = 31091.2 \text{ J} \). This heat absorption causes the helium gas to expand and affects the internal energy of the system.

The internal energy change \( \Delta E \) is crucial for understanding how energy transitions within a system. This change is calculated from the first law of thermodynamics, where we found \( \Delta E \) by summing heat absorbed \( q \) and work done \( w \).
In our example,

• Heat added, \( q = 31091.2 \text{ J} \)
• Work done, \( w = -12361.9 \text{ J} \)
• Thus, \( \Delta E = 31091.2 - 12361.9 = 18729.3 \text{ J} \)

This calculation shows that while the system absorbs a significant amount of energy as heat, some energy is expended as work. Therefore, the overall increase in internal energy reflects both heat absorption and energy spent in doing work. Understanding \( \Delta E \) is pivotal because it gives insight into the net energy change impacting the system's state.

The calculation of work done by the gas during expansion is a vital component of thermodynamics, especially in systems where volume change occurs at constant pressure. The formula for work \( w \) is:
\[ w = -P \Delta V \]
where \( P \) is pressure and \( \Delta V \) is the change in volume. In our case:

• Pressure \( P = 1 \text{ atm} \), converted to \( 101325 \text{ Pa} \)
• Initial volume \( V_{\text{initial}} = 876 \text{ L} \)
• Final volume \( V_{\text{final}} = 998 \text{ L} \) (converted to cubic meters, \( 0.122 \text{ m}^3 \))

Using these values in the equation gives the work as \( -12361.9 \text{ J} \), indicating energy is released to do the work of expansion. The negative sign shows that the gas does work on the surroundings, illustrating energy loss from the system as it pushes against atmospheric pressure to expand in volume.