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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 \(3.00 \times 10^{5} \mathrm{Pa}\) . The total heat liberated by the gas is \(2.50 \times 10^{4} \mathrm{J}\) . 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?

Short Answer

Expert verified
(a) 37.10 mol, (b) -24,838 J, (c) -1.23 x 10³ J, (d) -4.30 x 10³ J

Step by step solution

01

Convert Temperatures

Convert the temperatures from Celsius to Kelvin.For initial temperature: \[ T_1 = 50.0^{\circ}C + 273.15 = 323.15 \, K \]For final temperature:\[ T_2 = 10.0^{\circ}C + 273.15 = 283.15 \, K \]
02

Find the Number of Moles (Part a)

Use the ideal gas law in terms of initial and final conditions to find the number of moles. Since the pressure is constant, we can use the relation:\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]and the equation for work done at constant pressure: \[ W = P(V_2 - V_1) \]But first, solve for \( n \) using \[ PV = nRT \]Substitute known values\[ P = 3.00 \times 10^5 \, \text{Pa}, \]\[ T_1 = 323.15 \, K, \]\[ V_2/V_1 = T_2/T_1 \]Assuming \( V_1 = nRT_1/P \), then:\[ n = \frac{P \cdot V_1}{R \cdot T_1} \] \[ = \frac{3.00 \times 10^5}{8.314 \times 323.15} \approx 37.10 \, \text{mol} \]
03

Determine the Change in Internal Energy (Part b)

Use the first law of thermodynamics to relate heat, work, and internal energy:\[ \Delta U = Q - W \]Given that:- Heat liberated \( Q = -2.50 \times 10^4 \, J \)- Work \( W = P(V_2 - V_1) = 3.00 \times 10^5 (V_2 - V_1) \)Since \( V_2 / V_1 = T_2 / T_1 \), we calculate\[ \Delta U = -2.50 \times 10^4 - P \left( 3.00 \times 10^5 \right) \left(\frac{T_2}{T_1} -1 \right) nRT/3.00 \times 10^5 = -2.50 \times 10^4 \approx -24,838 \, J \]
04

Calculate the Work Done (Part c)

Again, we use \[ W = P(V_2 - V_1) \]\[ = 3.00 \times 10^5 (\frac{nRT_2/P - nRT_1/P}) \times (323.15 - 283.15) = -1.23 \times 10^3 \, J \]
05

Heat at Constant Volume (Part d)

Use the heat capacity at constant volume, where \[ Q = nC_v \Delta T \]For nitrogen,\[ C_v \text{ is roughly 29 J/mol K} \]Calculate:\[ Q = 37.10 \, \text{mol} \times 29 \, \text{J/mol K} \times (283.15 - 323.15) \approx -4.30 \times 10^3 \, J \]

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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 foundational principle in physics and chemistry. It states that energy cannot be created or destroyed, only transferred or converted from one form to another. In the context of an ideal gas, this law can be expressed as: \[ \Delta U = Q - W \] where:
  • \( \Delta U \) is the change in the internal energy of the system.
  • \( Q \) is the heat added to the system (negative if heat is removed).
  • \( W \) is the work done by the system (positive when the system does work on the surroundings).
For the process in the exercise, heat is liberated by the gas, meaning \( Q \) is negative. Therefore, to find the internal energy change, you must subtract the work done by the gas from the negative heat. This concept illustrates the conservation of energy by balancing inflow and outflow within the gas system.
Heat Capacity
Heat capacity is an important concept that represents how much heat is required to change the temperature of a substance by a certain amount. It depends on whether the gas is at constant volume or constant pressure:
  • Constant Volume (\( C_v \)): No expansion occurs, and it reflects the energy required without any work being done by the gas.
  • Constant Pressure (\( C_p \)): The gas can expand, and it accounts for both temperature change and work done during expansion.
In this text's scenario, heat is liberated when cooling nitrogen with a constant volume. Therefore, the heat capacity at constant volume \( C_v \) is used. For nitrogen gas, \( C_v \) is approximately 29 J/mol K. It's pivotal in calculating the heat exchange when the volume does not change as it directly relates the temperature change and the heat involved using \( Q = nC_v\Delta T \).
Internal Energy
Internal energy \( U \) of a gas is the total energy contained within it. For ideal gases, this energy comes primarily from the kinetic energy of molecules. Internal energy in a gas depends on the number of moles \( n \) and the temperature \( T \), following \[ U = \frac{3}{2}nRT \] where \( R \) is the ideal gas constant.In practical applications, the change in internal energy \( \Delta U \) is often of interest, especially when relating to processes such as cooling or heating like in this exercise, where the gas cools from 50.0°C to 10.0°C. Because temperature drops, there's a loss in internal energy which can be determined using the expression derived from the First Law of Thermodynamics: \[ \Delta U = Q - W \].This change tells us how much less kinetic energy is stored in the gas particles due to the cooling.
Work Done by Gas
The work done by or on a gas is a key concept that helps in understanding energy exchanges during expansion or compression. The work \( W \) done by the gas is calculated for processes at constant pressure using the formula:\[ W = P \Delta V \] where:
  • \( P \) is the constant pressure exerted on or by the gas.
  • \( \Delta V \) is the change in volume (difference between final and initial volume).
In scenarios where the volume change is related to temperature change, another form of the equation can be used based on the ideal gas law, substituting \( nRT \) for \( PV \). In this exercise, the gas does work as it cools and its volume changes due to temperature decrease. Calculating this work involves understanding the ideal gas relations between temperature, pressure, and volume, combining them to find the extent of energy transferred as mechanical work.

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Most popular questions from this chapter

In a cylinder sealed with a piston, you rapidly compress 3.00 \(\mathrm{L}\) of \(\mathrm{N}_{2}\) gas initially at 1.00 atm pressure and \(0.00^{\circ} \mathrm{C}\) to half its original volume. Assume the \(\mathrm{N}_{2}\) behaves like an ideal gas. (a) Calculate the final temperature and pressure of the gas. (b) If you now cool the gas back to \(0.00^{\circ} \mathrm{C}\) without changing the pressure, what is its final volume?

High-Altitude Research. A large research balloon containing \(2.00 \times 10^{3} \mathrm{m}^{3}\) of helium gas at 1.00 atm and a temperature of \(15.0^{\circ} \mathrm{C}\) rises rapidly from ground level to an altitude at which the atmospheric pressure is only 0.900 atm \((\text { Fig. } 19.33)\) . Assume the helium behaves like an ideal gas and the balloon's ascent is too rapid to permit much heat exchange with the surrounding air. (a) Calculate the volume of the gas at the higher altitude. (b) calculate the temperature of the gas at the higher altitude. (c) What is the change in internal energy of the helium as the balloon rises to the higher altitude?

An ideal gas expands while the pressure is kept constant. During this process, does heat flow into the gas or out of the gas? Justify your answer.

Chinook. During certain seasons strong winds called chinooks blow from the west across the eastern slopes of the Rockies and downhill into Denver and nearby areas. Although the mountains are cool, the wind in Denver is very hot; within a few minutes after the chinook wind arrives, the temperature can climb 20 \(\mathrm{C}^{\circ}\) ("chinook" is a Native American word meaning "snow cater"). Similar winds occur in the Alps (called fochns) and in southern California (called Santa Anas). (a) Explain why the temperature of the chinook wind rises as it descends the slopes. Why is it important that the wind be fast moving? (b) Suppose a strong wind is blowing toward Denver (elevation 1630 \(\mathrm{m} )\) from Grays Peak \((80 \mathrm{km} \text { west of Denver, at an elevation of } 4350 \mathrm{m})\) , where the air pressure is \(5.60 \times 10^{4} \mathrm{Pa}\) and the air temperature is \(-15.0 ^{\circ} \mathrm{C}\) . The temperature and pressure in Denver before the wind arrives are \(20^{\circ} \mathrm{C}\) and \(8.12 \times 10^{4} \mathrm{Pa}\) . By how many Celsius degrees will the temperature in Denver rise when the chinook arrives?

A cylinder with a movable piston contains 3.00 \(\mathrm{mol}\) of \(\mathrm{N}_{2}\) gas (assumed to behave like an ideal gas). (a) The \(\mathrm{N}_{2}\) is heated at constant volume until 1557 \(\mathrm{J}\) of heat have been added. Calculate the change in temperature. (b) Suppose the same amount of heat is added to the \(\mathrm{N}_{2}\) , but this time the gas is allowed to expand while remaining at constant pressure. Calculate the temperature change. (c) In which case, (a) or \((b),\) is the final internal energy of the \(\mathbf{N}_{2}\) higher? How do you know? What accounts for the difference between the two cases?

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