Problem 39 - Chemistry An Atoms First Approach

Question

A balloon filled with \(39.1 \mathrm{moles}\) of helium has a volume of \(876 \mathrm{L}\) at \(0.0^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm}\) pressure. The temperature of the balloon is increased to \(38.0^{\circ} \mathrm{C}\) as it expands to a volume of \(998 \mathrm{L}\), the pressure remaining constant. Calculate \(q, w,\) and \(\Delta E\) for the helium in the balloon. (The molar heat capacity for helium gas is \(20.8 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{mol}.\))

Step-by-Step Solution

Answer

The heat (q) absorbed by the helium gas is \(30872.56 \mathrm{J}\), the work (w) done by the gas is \(-12361.7 \mathrm{J}\), and the change in internal energy (ΔE) of the gas is \(18510.86 \mathrm{J}\).

Step 1: Convert temperatures to Kelvin

Since we will be working with the Ideal Gas Law and temperature changes, it's important to convert the given temperatures from Celsius to Kelvin. Initial temperature, \(T_1 = 0.0^{\circ} \mathrm{C} = 273.15 \mathrm{K}\) Final temperature, \(T_2 = 38.0^{\circ} \mathrm{C} = 273.15 + 38.0 = 311.15 \mathrm{K}\)

Step 2: Calculate the change in volume

To find the work done by the helium gas, we need to find the change in volume. ΔV = final volume - initial volume ΔV = \(998 \mathrm{L} - 876 \mathrm{L} = 122 \mathrm{L}\)

Step 3: Calculate the work done

We know the following equation for work done by a gas when the pressure is constant: w = -PΔV In this case, the constant pressure P is given as \(1.00 \mathrm{atm}\). Since we are given heat capacity in J/°C, we should change the pressure unit from atm to Pascal. \(1 \mathrm{atm} = 101325 \mathrm{Pa}\) So, w = \(-1.00 \times 101325 \mathrm{Pa} \times 122 \mathrm{L}\) Convert litres to cubic meters: \(1\mathrm{L} = 0.001\mathrm{m^3}\) w = \(-101325 \mathrm{Pa} \times 0.122\mathrm{m^3} = -12361.7 \mathrm{J}\) The work done by the helium gas is -12361.7 J.

Step 4: Calculate the heat absorbed by the gas

We can use the following equation for the heat absorbed for a constant pressure process: q = nCΔT Here, n = 39.1 moles, C = 20.8 J/mol°C, and ΔT = T2 - T1 = 311.15 K - 273.15 K = 38.0 K. So, q = \(39.1 \mathrm{mol} \times 20.8 \frac{\mathrm{J}}{\mathrm{mol\cdot K}} \times 38.0\mathrm{K} = 30872.56\mathrm{J}\) The heat absorbed by the helium gas is 30872.56 J.

Step 5: Calculate the change in internal energy

Now that we have both q and w, we can use the first law of thermodynamics to find ΔE: ΔE = q + w ΔE = \(30872.56 \mathrm{J} - 12361.7 \mathrm{J} = 18510.86 \mathrm{J}\) The change in internal energy of the helium gas is 18510.86 J. Solution: The heat (q) absorbed by the helium gas is 30872.56 J, the work (w) done by the gas is -12361.7 J, and the change in internal energy (ΔE) of the gas is 18510.86 J.

Key Concepts

Ideal Gas Law

The ideal gas law is a cornerstone of thermodynamics, providing a relationship between the pressure, volume, and temperature of an ideal gas. This law is expressed by the equation \( PV = nRT \), where:

• \( P \) represents the pressure of the gas.
• \( V \) indicates the volume of the gas.
• \( n \) is the number of moles of the gas.
• \( R \) is the universal gas constant \( (8.314 \,\mathrm{J\,mol^{-1}\,K^{-1}}) \).
• \( T \) is the temperature in Kelvin.

The ideal gas law helps us understand how adjustments in temperature and pressure can affect the volume of a gas. In the given problem, converting temperatures from Celsius to Kelvin is crucial, as the gas law equation requires temperatures in Kelvin. This conversion is simply done by adding 273.15 to the temperature in Celsius. Knowing that the pressure is constant allows us to focus on changes in volume and temperature to calculate other thermodynamic properties.

Molar Heat Capacity

Molar heat capacity is a measure of the amount of heat required to increase the temperature of one mole of a substance by one degree Celsius (or one Kelvin). It is denoted by \( C \) and is expressed in units of \( \mathrm{J/mol\, °C} \) or \( \mathrm{J/mol\, K} \).
The molar heat capacity at constant pressure \( C_p \) is especially important when dealing with gases like helium in our example. Here, the molar heat capacity of helium gas is given as \( 20.8 \,\mathrm{J/mol\,°C} \).
Using the relationship \( q = nC_p \Delta T \), where:

• \( q \) is the heat absorbed or released.
• \( n \) indicates the number of moles.
• \( \Delta T \) is the change in temperature, obtained by subtracting the initial temperature from the final temperature (in Kelvin).

We can calculate the amount of heat involved in the process for the helium in the balloon. This calculation is fundamental to determining how energy is distributed within the system during temperature changes.

Internal Energy

Internal energy (\( \Delta E \)) refers to the total energy contained within a thermodynamic system. It represents the sum of the potential and kinetic energy of the molecules in the system. Changes in internal energy correspond to the energy added or removed from the system, either by heat transfer or by doing work.
The first law of thermodynamics is key to understanding internal energy changes and is stated as \( \Delta E = q + w \). Here:

• \( q \) is the heat exchanged with the surroundings.
• \( w \) is the work done by or on the system.

For the helium in the balloon, after calculating the work done and the heat absorbed, we can easily compute the change in internal energy using this equation. Understanding this concept helps explain how energy conservation works in thermodynamic processes. The calculation reveals how heat absorbed corresponds with energy changes internally, without violating conservation rules.

Work Done by Gas

In thermodynamics, work done by a gas during its expansion or compression is an important concept. When a gas does work, its internal energy changes, often along with other state variables such as volume, pressure, and temperature. The work done by a gas is calculated using the equation \( w = -P \Delta V \), especially in constant pressure scenarios.

• \( P \) is the pressure of the gas (constant in this problem).
• \( \Delta V \) is the change in volume, the difference between the final and initial volumes.

For the given balloon problem, converting pressure from atmospheres (atm) to Pascals (Pa) is necessary to maintain consistent units. As gases expand, they often perform work on the surroundings by displacing volume. Negative signs indicate that work is done by the gas when it expands. Calculating this work helps in understanding how systems can exchange energy with their environment, contributing to changes in internal energy.