Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Nitrogen gas in an expandable container is cooled from 50.0\(^\circ\)C to 10.0\(^\circ\)C with the pressure held constant at 3.00 \(\times\) 10\(^5\) Pa. The total heat liberated by the gas is 2.50 \(\times\) 10\(^4\) J. Assume that the gas may be treated as ideal. Find (a) the number of moles of gas; (b) the change in internal energy of the gas; (c) 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) Use ideal gas laws to evaluate mole values. b) Compute \( \Delta U \) from heat and work changes. c) Apply \( W = nR\Delta T \), verifying contracted volumes. d) Utilize \( Q_v \) measurement neglects work via molar heat roles.

Step by step solution

01

Convert Temperatures to Kelvin

Start by converting the given temperatures from degrees Celsius to Kelvin. The conversion formula is \( T(K) = T(^\circ C) + 273.15 \).\(T_1 = 50.0 + 273.15 = 323.15\, K \) \(T_2 = 10.0 + 273.15 = 283.15\, K \)
02

Calculate Number of Moles (Part a)

Use the ideal gas law \( PV = nRT \) to solve for the number of moles \( n \). Assume the same \( V \) for initial and final conditions due to pressure held constant. Since \( V \) does not change, calculate \( n \) using either condition, for example: \( n = \frac{PV}{RT_1}\)1) Substitute the ideal gas constant \( R = 8.314 \) J/(mol·K): \( n = \frac{3.00 \times 10^5 \times V}{8.314 \times 323.15}\)Here, \( V \) cancels out, and you can rearrange in terms of other known quantities if needed.
03

Find Change in Internal Energy (Part b)

Use the first law of thermodynamics, \( \Delta U = Q - W \), where \(\Delta U \) is the change in internal energy, \( Q \) is heat change, and \( W \) is work done by the gas. When volume is not constant, solve for \( W \) first via \( W = P\Delta V \), requiring next actual volume calculations if not already asserted constant. With heat liberated, calculate \( \Delta U \): \( \Delta U = Q - W \) Given \( Q = -2.50 \times 10^4 \) J (indicating output), further evaluate with \( W \) understanding.
04

Calculate Work Done by the Gas (Part c)

Since pressure is constant and volume changes, let's solve for work, \( W = P\Delta V \). Use \(\Delta V = V_2 - V_1 \). Considering the entire reversible process: \( W = nR(T_2 - T_1) \) derived with simplicity is sufficient noting equalities. Given work should be negative indicating cooling, leading to contraction:
05

Calculate Heat at Constant Volume (Part d)

For constant volume conditions, use: \( Q_v = nC_v \Delta T \) where \( C_v = \frac{f}{2}R \), the constant volume molar heat capacity for diatomic molecules, \( f = 5 \).Replace values for nitrogen: - Compute \( \Delta T = T_2 - T_1 \) using Kelvin: \( Q_v = n \times \left(\frac{5}{2} \times 8.314\right) \times (283.15 - 323.15) \). Recall \( \Delta U \) can also apply due lack of exertion.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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 lays the groundwork for understanding energy interactions in physical systems. It's essentially about the conservation of energy, stating that energy cannot be created or destroyed, only transformed or transferred. This law is expressed as: \[ \Delta U = Q - W \]where \( \Delta U \) represents the change in internal energy of the system, \( Q \) is the heat added to the system, and \( W \) is the work done by the system.
In the context of the exercise, as the nitrogen gas cools and contracts under constant pressure, it does work on the surroundings by reducing its volume. The release of heat (\( Q = -2.50 \times 10^4 \) J) results in a decrease in the internal energy of the gas, adjusted by the work term \( W \) due to the change in volume.
  • Heat \( Q \) is negative here since the gas is losing heat.
  • Work \( W \) signals changes in volume under constant pressure, impacting \( \Delta U \).
Understanding this relationship helps elucidate how transferring energy as heat and work affects a gas's internal state.
Heat Transfer
Heat transfer deals with the movement of thermal energy from a region of higher temperature to a region of lower temperature. In the given exercise, the nitrogen gas undergoes a decrease in temperature, with heat being liberated as it is removed from the gas to its surroundings.
Heat transfer can occur in three ways: conduction, convection, and radiation, but here it is considered in relation to thermodynamic processes happening at constant pressure.
  • The process is governed by the ideal gas law, where temperature changes induce volume changes.
  • The amount of heat transfer is dictated by the initial and final states of the temperature.
The exercise part (d) adds clarity to how the gas behaves if the volume is held constant. Differentiating between constant volume and constant pressure conditions is crucial. At constant volume, the full heat exchange is used to change the internal energy without performing any work (since \( W = 0 \)).
Molar Heat Capacity
Molar heat capacity is a measure of the amount of heat required to raise the temperature of one mole of a substance by one degree Kelvin. For gases, this concept varies significantly depending on whether the volume or pressure is constant.
When dealing with diatomic gases like nitrogen, different constants are used:
  • \( C_p \) (heat capacity at constant pressure): How much heat is needed per mole to heat the gas while allowing it to expand.
  • \( C_v \) (heat capacity at constant volume): How much heat is needed when the gas is confined to a fixed volume.
For part (d) of the exercise, it's calculated under constant volume conditions using \( C_v = \frac{f}{2}R \), where \( f = 5 \) for diatomic nitrogen. This indicates that less heat would be liberated with no work done, as all energy change caters to internal energy alterations without any mechanical work exerted.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The power output of an automobile engine is directly proportional to the mass of air that can be forced into the volume of the engine's cylinders to react chemically with gasoline. Many cars have a \(turbocharger\), which compresses the air before it enters the engine, giving a greater mass of air per volume. This rapid, essentially adiabatic compression also heats the air. To compress it further, the air then passes through an \(intercooler\) in which the air exchanges heat with its surroundings at essentially constant pressure. The air is then drawn into the cylinders. In a typical installation, air is taken into the turbocharger at atmospheric pressure (1.01 \(\times\) 10\(^5\) Pa), density \(\rho\) = 1.23 kg/m\(^3\), and temperature 15.0\(^\circ\)C. It is compressed adiabatically to 1.45 \(\times\) 10\(^5\) Pa. In the intercooler, the air is cooled to the original temperature of 15.0\(^\circ\)C at a constant pressure of 1.45 \(\times\) 10\(^5\) Pa. (a) Draw a \(pV\)-diagram for this sequence of processes. (b) If the volume of one of the engine's cylinders is 575 cm\(^3\), what mass of air exiting from the intercooler will fill the cylinder at 1.45 \(\times\) 10\(^5\) Pa? Compared to the power output of an engine that takes in air at 1.01 \(\times\) 10\(^5\) Pa at 15.0\(^\circ\)C, what percentage increase in power is obtained by using the turbocharger and intercooler? (c) If the intercooler is not used, what mass of air exiting from the turbocharger will fill the cylinder at 1.45 \(\times\) 10\(^5\) Pa? Compared to the power output of an engine that takes in air at 1.01 \(\times\) 10\(^5\) Pa at 15.0\(^\circ\)C, what percentage increase in power is obtained by using the turbocharger alone?

During an isothermal compression of an ideal gas, 410 J of heat must be removed from the gas to maintain constant temperature. How much work is done by the gas during the process?

When water is boiled at a pressure of 2.00 atm, the heat of vaporization is 2.20 \(\times\) 10\(^6\) J/kg and the boiling point is 120\(^\circ\)C. At this pressure, 1.00 kg of water has a volume of 1.00 \(\times\) 10\(^{-3}\) m\(^3\), and 1.00 kg of steam has a volume of 0.824 m\(^3\). (a) Compute the work done when 1.00 kg of steam is formed at this temperature. (b) Compute the increase in internal energy of the water.

A cylinder with a frictionless, movable piston like that shown in Fig. 19.5 contains a quantity of helium gas. Initially the gas is at 1.00 \(\times\) 10\(^5\) Pa and 300 K and occupies a volume of 1.50 L. The gas then undergoes two processes. In the first, the gas is heated and the piston is allowed to move to keep the temperature at 300 K. This continues until the pressure reaches 2.50 \(\times\) 10\(^4\) Pa. In the second process, the gas is compressed at constant pressure until it returns to its original volume of 1.50 L. Assume that the gas may be treated as ideal. (a) In a \(pV\)-diagram, show both processes. (b) Find the volume of the gas at the end of the first process, and the pressure and temperature at the end of the second process. (c) Find the total work done by the gas during both processes. (d) What would you have to do to the gas to return it to its original pressure and temperature?

In an experiment to simulate conditions inside an automobile engine, 0.185 mol of air at 780 K and 3.00 \(\times\) 10\(^6\) Pa is contained in a cylinder of volume 40.0 cm\(^3\). Then 645 J of heat is transferred to the cylinder. (a) If the volume of the cylinder is constant while the heat is added, what is the final temperature of the air? Assume that the air is essentially nitrogen gas, and use the data in Table 19.1 even though the pressure is not low. Draw a \(pV\)-diagram for this process. (b) If instead the volume of the cylinder is allowed to increase while the pressure remains constant, repeat part (a).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free