Problem 52 - University Physics with Modern Physics

Question

Nitrogen gas in an expandable container is cooled from \(50.0^{\circ} \mathrm{C}\) to \(10.0^{\circ} \mathrm{C}\) with the pressure held constant at \(3.00 \times 10^{3}\) Pa. The total heat liberated by the gas is \(2.50 \times 10^{4}\) . Assume that the gas may be treated as ideal. (a) Find the number of moles of gas. (b) Find the change in internal energy of the gas. (c) Find the work done by the gas. (d) How much heat would be liberated by the gas for the same temperature change if the volume were constant?

Step-by-Step Solution

Answer

(a) 21.45 mol (b) -17,861.58 J (c) 42,861.58 J (d) -17,861.58 J

Step 1: Determine the temperature change

Convert the temperature from Celsius to Kelvin.\[T_1 = 50.0^{\circ}C + 273.15 = 323.15 \, K\]\[T_2 = 10.0^{\circ}C + 273.15 = 283.15 \, K\]Calculate the change in temperature:\[\Delta T = T_2 - T_1 = 283.15 - 323.15 = -40 \, K\]

Step 2: Calculate the number of moles using the ideal gas law

Since the pressure is constant, use the relationship from the ideal gas law:\[PV = nRT\]Between two states and at constant pressure:\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]From this, assuming constant amounts and rearranging for n:\[n = \frac{P(V)}{R(T)}\] We use:\[Q = nC_p\Delta T\]Plug into the derived formula:\[n = \frac{Q}{C_p\Delta T}\]Given that \(C_p = 29.1 \, J/mol \, K \) for nitrogen,\[n = \frac{2.50 \times 10^4 \, J}{29.1 \, J/mol \, K \times (-40 \, K)} = -21.45 \, mol\]Since only magnitude matters, \(n = 21.45 \, \,mol\).

Step 3: Calculate the change in internal energy

The formula for the change in internal energy \(\Delta U\) of an ideal gas is:\[\Delta U = nC_v\Delta T\]where \(C_v = C_p - R = 29.1 \, J/mol \, K - 8.31 \, J/mol \, K = 20.79 \, J/mol \, K\).Substitute to find \(\Delta U\):\[\Delta U = 21.45 \, mol \times 20.79 \, J/mol \, K \times (-40) \, K\]\[\Delta U = -17,861.58 \, J\].

Step 4: Calculate the work done by the gas

The work done by the gas at constant pressure is given by:\[W = P \Delta V\].Use the relation \(\Delta U = Q - W\) to solve for \(W\):\[W = Q - \Delta U = 2.50 \times 10^4 \, J + 17,861.58 \, J\]\[W = 42,861.58 \, J\].

Step 5: Calculate heat liberated at constant volume

At constant volume, the change in heat is equivalent to the change in internal energy.\[Q_{\text{constant V}} = \Delta U\]So, \[Q_{\text{constant V}} = -17,861.58 \, J\].

Key Concepts

Thermodynamics

Thermodynamics is the study of heat, energy, and the work that can be extracted or put into a physical system. It focuses on the principles that dictate how energy is transferred and transformed.
In particular, the Ideal Gas Law is a fundamental equation in thermodynamics. It describes how the pressure, volume, and temperature of an ideal gas relate to one another, expressed as \(PV = nRT\), where:

• \(P\) is the pressure of the gas.
• \(V\) is the volume.
• \(n\) is the number of moles.
• \(R\) is the ideal gas constant.
• \(T\) is the temperature in Kelvin.

In this exercise, nitrogen gas serves as an example of an ideal gas. Understanding how these properties change with constant pressure, as the gas is cooled, gives insight into the broader principles of thermodynamics.

Internal Energy

Internal energy is a vital concept in thermodynamics. It refers to the total energy contained within a system. For an ideal gas, the internal energy \( U \) depends entirely on its temperature. As the temperature changes, so does the internal energy.
In our problem, the change in the internal energy, denoted as \( \Delta U \, \), happens as the gas cools from a higher to a lower temperature. This is calculated using:

• \( \Delta U = nC_v\Delta T \)

where \( C_v \) is the specific heat capacity at constant volume. For nitrogen, \( C_v = 20.79 \, J/mol \, K \) as calculated from \( C_p - R \,\), indicating that the gas’s internal energy decreases as temperature drops.
This decrease in internal energy translates to heat being removed from the system, emphasizing the strong connection between temperature and internal energy in gases.

Work Done

Work done by or on a system is another key piece in thermodynamics. At constant pressure, work done by a gas is expressed as \( W = P \Delta V \). This relates to the energy required for a gas to either expand or compress.
In the exercise, the work done by the nitrogen gas can be determined indirectly by using the first law of thermodynamics:

• \( \Delta U = Q - W \)

Where \(Q\) is the heat exchanged. Solving for \(W\) gives us the work done when the gas expands as it cools and releases heat. The additional energy balance helps illustrate how thermodynamic systems can do work as they transfer energy through heat interactions.
Gas doing work while cooling, underlines not just how energy conservation holds, but also highlights the interplay between heat exchange, internal energy change, and mechanical work in a dynamic process.