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Chapter 19: Problem 28

Three moles of an ideal monatomic gas expands at a con-stant pressure of 2.50 atm; the volume of the gas changes from \(3.20 \times 10^{-2} \mathrm{m}^{3}\) to \(4.50 \times 10^{-2} \mathrm{m}^{3} .\) (a) Calculate the initial and final temperatures of the gas. (b) Calculate the amount of work the gas does in expanding. (c) Calculate the amount of heat added to the gas. (d) Calculate the change in internal encrgy of the gas.

Short Answer

Expert verified
(a) Initial: ~325.3 K, Final: ~457.1 K; (b) Work: 3296 J; (c) Heat: 214720 J; (d) Energy Change: 154720 J.

Step by step solution

01

Convert Pressure to SI Units

First, convert the pressure from atmospheres to Pascals for use in calculations. Given that \(1 \text{ atm} = 101325 \text{ Pa}\), we multiply: \(2.50 \text{ atm} \times 101325 \text{ Pa/atm} = 253312.5 \text{ Pa} \approx 2.53 \times 10^5 \text{ Pa}\).
02

Calculate Initial and Final Temperature

Use the ideal gas law \(PV = nRT\) to find the initial and final temperatures. **Initial Temperature:**\[P V_i = n R T_i\]\[T_i = \frac{P V_i}{n R} = \frac{(2.53 \times 10^5 \text{ Pa})(3.20 \times 10^{-2} \text{ m}^3)}{3 \times 8.314 \text{ J/mol K}} \approx 325.3 \text{ K}\]**Final Temperature:**\[P V_f = n R T_f\]\[T_f = \frac{P V_f}{n R} = \frac{(2.53 \times 10^5 \text{ Pa})(4.50 \times 10^{-2} \text{ m}^3)}{3 \times 8.314 \text{ J/mol K}} \approx 457.1 \text{ K}\]
03

Calculate Work Done by the Gas

The work done by a gas at constant pressure during expansion is given by \(W = P\Delta V\), where \(\Delta V = V_f - V_i\).\[W = 2.53 \times 10^5 \text{ Pa} \times (4.50 \times 10^{-2} \text{ m}^3 - 3.20 \times 10^{-2} \text{ m}^3) = 3296.25 \text{ J}\]
04

Calculate Heat Added to the Gas

For a monatomic ideal gas, the heat added can be found using \(Q = nC_p\Delta T\) where \(C_p = \frac{5}{2}R\).\[Q = 3 \times \frac{5}{2} \times 8.314 \times (457.1 - 325.3)\]\[Q \approx 1630.89 \times 131.8 = 214720.3 \text{ J}\]
05

Calculate Change in Internal Energy

The change in internal energy \(\Delta U\) for a monatomic ideal gas can be calculated using \(\Delta U = nC_v\Delta T\) where \(C_v = \frac{3}{2}R\).\[\Delta U = 3 \times \frac{3}{2} \times 8.314 \times (457.1 - 325.3)\]\[\Delta U \approx 1171.41 \times 131.8 = 154720.2 \text{ J}\]
06

Validate Energy Conservation

Check that the first law of thermodynamics \(\Delta U = Q - W\) holds:\[154720.2 \text{ J} \approx 214720.3 \text{ J} - 3296.25 \text{ J}\] Verify both sides of the equation are equal within reasonable rounding boundaries.

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Key Concepts

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

Thermodynamics
Thermodynamics is the science related to the study of energy interactions, primarily focusing on heat and work transfers. In our scenario, we are dealing with an ideal monatomic gas undergoing expansion at constant pressure. This context involves understanding how energy is transferred and transformed in the system.
A key aspect of thermodynamics is the ideal gas law, which states that \(PV = nRT\), where:
  • \(P\) is pressure in Pascals,
  • \(V\) is volume in cubic meters,
  • \(n\) is the number of moles,
  • \(R\) is the universal gas constant (8.314 J/mol K), and
  • \(T\) is the temperature in Kelvin.
The ideal gas law will help us compute the initial and final temperatures when the gas expands. Understanding these basic principles is crucial to solving problems involving gases under different conditions.
Work Done by Gas
When a gas expands, it can do work on its surroundings. For a process at constant pressure, like in this exercise, the work done by a gas can be calculated with the formula \(W = P \Delta V\), where \(\Delta V = V_f - V_i\) is the change in volume.
This equation tells us that:
  • The work done (\(W\)) is positive when a gas expands because it pushes against the surrounding pressure.
  • \(P\) is the constant external pressure applied to the system.
  • The greater the change in volume, the more work is done.
In our specific case, knowing the initial and final volumes, along with the constant pressure, allows us to calculate the total work done during the gas expansion.
Change in Internal Energy
Internal energy (\(U\)) of a system is associated with the total kinetic energy possessed by the molecules of the system. For a monatomic ideal gas, the change in internal energy (\(\Delta U\)) is determined by the formula \(\Delta U = nC_v \Delta T\).
The key points are:
  • \(C_v = \frac{3}{2}R\) is the specific heat at constant volume for a monatomic gas.
  • \(\Delta T = T_f - T_i\) represents the change in temperature.
This relationship helps us see that any change in temperature directly affects the internal energy. Internal energy increases with an increase in temperature, meaning the gas molecules move faster and possess more kinetic energy.
Heat Transfer
In thermodynamics, heat transfer refers to the movement of thermal energy from one place to another. Using the equation \(Q = nC_p \Delta T\) for this solution, we assess the heat added to our monatomic ideal gas.
Here:
  • \(Q\) is the heat added to the system,
  • \(C_p = \frac{5}{2}R\) is the specific heat capacity at constant pressure for a monatomic gas, and
  • \(\Delta T = T_f - T_i\) is the change in temperature.
The first law of thermodynamics states that the change in internal energy \(\Delta U\) is equal to the heat added to the system minus the work done by the system: \(\Delta U = Q - W\). Understanding this balance helps us evaluate how energy flows through the system, ensuring conservation of energy within the framework of thermodynamics.

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Most popular questions from this chapter

On a warm summer day, a large mass of air (armospheric pressure \(1.01 \times 10^{5} \mathrm{Pa}\) ) is heated by the ground to a temperature of \(26.0^{\circ} \mathrm{C}\) and then begins to rise through the cooler surrounding air. (This can be treated approximately as an adiabatic process; why? Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only \(0.850 \times 10^{5} \mathrm{Pa}\) . Assume that air is an ideal gas, with \(\gamma=1.40\) . (This rate of cool- ing for dry, rising air, corresponding to roughly \(1^{\circ} \mathrm{C}\) per 100 \(\mathrm{m}\) of attitude, is called the dry adiabatic lapse rate.)

A gas in a cylinder expands from a volume of 0.110 \(\mathrm{m}^{3}\) to 0.320 \(\mathrm{m}^{3}\) . Heat flows into the gas just rapidly enough to keep the pressure constant at \(1.80 \times 10^{5}\) Pa during the expansion. The total heat added is \(1.15 \times 10^{5} \mathrm{J}\) . (a) Find the work done by the gas. (b) Find the change in internal energy of the gas. it matter whether the gas is ideal? Why or why not?

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?

Oscillations of a Piston. A vertical cylinder of radins \(r\) contains a quantity of ideal gas and is fitted with a piston with mass \(m\) that is free to move (Fig. 19.34\()\) . The piston and the walls of the cylinder are frictionless and the entire cylinder is placed in aconstant-temperature bath. The outside air pressure is \(p_{0}\) . In equilibrium, the piston sits at a height \(h\) above the bottom of the cylinder. (a) Find the absolute pressure of the gas trapped below the piston when in equilibrium. (b) The piston is pulled up by a small distance and released. Find the net force acting on the piston when its base is a distance \(h+y\) above the bottom of the cylinder, where \(y\) is much less than \(h\) (c) After the piston is displaced from equilibrium and released, it oscillates up and down. Find the frequency of nthese small oscillations. If the displacement is not small, are the oscillations simple harmonic? How can you tell?
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