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The temperature of \(5.00 \mathrm{~kg}\) of \(\mathrm{N}_{2}\) gas is raised from \(10.0{ }^{\circ} \mathrm{C}\) to \(130.0{ }^{\circ} \mathrm{C}\). If this is done at constant volume, find the increase in internal energy \(\Delta U .\) Alternatively, if the same temperature change now occurs at constant pressure determine both \(\Delta V\) and the external work \(\Delta W\) done by the gas. For \(\mathrm{N}_{2}\) gas, \(c_{v}=0.177 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and \(c_{p}=0.248 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). If the gas is heated at constant volume, then no work is done during the process. In that case \(\Delta W=0\), and the first law tells us that \((\Delta Q)_{v}=\Delta U .\) Because \((\Delta Q)_{v}=c_{v} m \Delta T\), $$\Delta U=(\Delta Q)_{v}=\left(0.177 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(5000 \mathrm{~g})\left(120^{\circ} \mathrm{C}\right)=106 \mathrm{kcal}=443 \mathrm{~kJ}$$ The temperature change is a manifestation of the internal energy change. When the gas is heated \(120^{\circ} \mathrm{C}\) at constant pressure, the same change in internal energy occurs. In addition, however, work is done. The first law then becomes $$(\Delta Q)_{p}=\Delta U+\Delta W=443 \mathrm{~kJ}+\Delta W$$ But $$\begin{aligned} (\Delta Q)_{p} &=c_{p} m \Delta T=\left(0.248 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(5000 \mathrm{~g})\left(120{ }^{\circ} \mathrm{C}\right) \\ &=149 \mathrm{kcal}=623 \mathrm{~kJ} \end{aligned}$$ $$ \text { Hence } \quad \Delta W=(\Delta Q)_{p}-\Delta U=623 \mathrm{~kJ}-443 \mathrm{~kJ}=180 \mathrm{~kJ} $$

Step by step solution

01

Calculate Change in Internal Energy at Constant Volume

When the nitrogen gas is heated at constant volume, no work is done (\(\Delta W = 0\)), and the change in internal energy is equivalent to the heat added \((\Delta Q)_v\). The formula is \(\Delta U = c_v \cdot m \cdot \Delta T\)., where \(c_v = 0.177 \text{ cal/g°C}, m = 5000 \text{ g}, \text{ and } \Delta T = 130.0°C - 10.0°C = 120.0°C\). Thus:\[\Delta U = \left(0.177 \text{ cal/g°C}\right) \times (5000 \text{ g}) \times (120.0°C) = 106 \text{ kcal}\]\Converting to joules, since \(1 \text{ kcal} = 4.184 \text{ kJ}\), this becomes:\[\Delta U = 106 \text{ kcal} \times 4.184 \text{ kJ/kcal} = 443 \text{ kJ}\]The change in internal energy at constant volume is \(443 \text{ kJ}\).

02

Calculate Total Heat Added at Constant Pressure

When the nitrogen gas is heated at constant pressure, the heat added \((\Delta Q)_p\), can be calculated as:\((\Delta Q)_p = c_p \cdot m \cdot \Delta T\), where \(c_p = 0.248 \text{ cal/g°C}, m = 5000 \text{ g}, \text{ and } \Delta T = 120.0°C\). Thus:\[(\Delta Q)_p = (0.248 \text{ cal/g°C}) \times (5000 \text{ g}) \times (120.0°C) = 149 \text{ kcal}\]Converting to joules:\[(\Delta Q)_p = 149 \text{ kcal} \times 4.184 \text{ kJ/kcal} = 623 \text{ kJ}\]The heat added at constant pressure is \(623 \text{ kJ}\).

03

Determine Work Done at Constant Pressure

According to the first law of thermodynamics, at constant pressure, the heat added \((\Delta Q)_p\) is equal to the change in internal energy plus the work done \((\Delta W)\) by the system: \((\Delta Q)_p = \Delta U + \Delta W\).We already know:\(\Delta Q_p = 623 \text{ kJ}\) and\(\Delta U = 443 \text{ kJ}\).Thus, rearranging the formula gives:\[\Delta W = (\Delta Q)_p - \Delta U = 623 \text{ kJ} - 443 \text{ kJ} = 180 \text{ kJ}\]The work done by the gas at constant pressure is \(180 \text{ kJ}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Law of Thermodynamics

The First Law of Thermodynamics is a fundamental principle that connects internal energy, heat, and work. It states that the energy of an isolated system remains constant. This means that energy can neither be created nor destroyed, only transformed or transferred. The formula often used is\[ \Delta U = \Delta Q - \Delta W \]where \( \Delta U \) is the change in internal energy, \( \Delta Q \) is the heat added to the system, and \( \Delta W \) is the work done by the system.

This law is essentially about energy conservation. For instance, in our exercise, when the gas is heated at constant volume, no work is done, so all heat energy goes into changing the internal energy. However, at constant pressure, part of the energy goes into doing work by expanding the gas.

Internal Energy

The internal energy of a system refers to the total energy contained within it, arising from the kinetic and potential energies of its molecules. This concept forms the foundation for understanding changes in temperature and phase.

When the gas was heated, the rise in temperature signified an increase in kinetic energy, thus increasing the internal energy. This energy change is calculated using the formula:\[ \Delta U = c_v \cdot m \cdot \Delta T \]where \( c_v \) is the specific heat capacity at constant volume, \( m \) is the mass, and \( \Delta T \) is the temperature change.

In our example, this helped us find that the internal energy change of the nitrogen gas was 443 kJ.

Heat Capacity

Heat capacity is an important property that describes how much heat energy is needed to change a substance's temperature. There are two types of specific heat capacity to consider: at constant volume \(c_v\) and at constant pressure \(c_p\).

: This describes the heat required to raise the temperature of a unit mass by one degree Celsius, without changing the volume. In the exercise, the value \(0.177 \text{ cal/g°C}\) was used for calculations involving constant volume.
: Conversely, this is the heat required at constant pressure, where the gas can expand or contract. For nitrogen gas, it was \(0.248 \text{ cal/g°C}\).

These values were crucial for figuring out both the internal energy change and the work done by the gas in different conditions.

Ideal Gas Law

The Ideal Gas Law is a simplified equation of state for a gas, expressed as:\[ PV = nRT \]where \(P\) is pressure, \(V\) is volume, \(n\) is the amount of substance in moles, \(R\) is the gas constant, and \(T\) is temperature.

While it wasn’t directly calculated here, this law helps to understand the behavior of gases under various conditions. For example, it can help predict how a gas's volume changes with pressure and temperature. During the constant pressure process from our exercise, the gas’s volume expanded as it heated, tying in with this concept. Ideal gases, like nitrogen, follow the law under a range of conditions, linking back to the calculations and explanations provided in the thermodynamics context.