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How would the temperature of a REAL gas (not perfect) in a reversible adiabatic expansion be affected?
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- Compare the charge in internal energy of an ideal gas for a quasi-static adiabatic expansion with that for a quasi-static isothermal expansion. What happens to the temperature of an ideal gas in an adiabatic expansion?An ideal diatomic gas at 80 K is slowly compressed adiabatically to one-third its original volume. What is its final temperature?Two moles of a monatomic ideal gas such as oxygen is compressed adiabatically and reversibly from a state (3 atm, 5 L) to a state with a pressure of 4 atm. (a) Find the volume and temperature of the final state. (b) Find the temperature of the initial state. (c) Find work done by the gas in the process. (d) Find the change in internal energy in the process. Assume Cv=5R and Cp=Cv+R for the diatomic ideal gas in the conditions given.
- When 400 J of heat are slowly added to 10 mol of an ideal monatomic gas, its temperature rises by 10 . What is the work done on the gas?An ideal monatomic gas at 300 K expands adiabatically and reversibly to twice its volume. What is its final temperature?A car tile contains 0.0380 m3 of air at a pressure of 2.20105 Pa (about 32 psi). How much more internal energy does this gas have than the same volume has at zero gauge pressure (which is equivalent to normal atmospheric pressure)?
- Two moles of nitrogen gas, with =7/5 for ideal diatomic gases, occupies a volume of 102 m3 in an insulated cylinder at temperature 300 K. The gas is adiabatically and reversibly compressed to a volume of 5 L. The piston of the cylinder is locked in its place, and the insulation around the cylinder is removed. The heat-conducting cylinder is then placed in a 300-K bath. Heat from the compressed gas leaves the gas, and the temperature of the gas becomes 300 K again. The gas is then slowly expanded at the fixed temperature 300 K until the volume of the gas becomes 102 m3, thus making a complete cycle for the gas. For the entire cycle, calculate (a) the work done by the gas, (b) the heat into or out of the gas, (c) the change in the internal energy of the gas, and (d) the change in entropy of the gas.An ideal gas has a pressure of 0.50 atm and a volume of 10 L. It is compressed adiabatically and quasi-statically until its pressure is 3.0 atm and its volume is 2.8 L. Is the monatomic, diatomic, or polyatomic?On an adiabatic process of an ideal gas pressure, volume and temperature change such that pV is constant with =5/3 for monatomic gas such as helium and =7/5 for diatomic gas such as hydrogen at room temperature. Use numerical values to plot two isotherms of 1 mol of helium gas using ideal gas law and two adiabatic processes mediating between them. Use T1=500K,V1=1L, and T2=300K for your plot.
- For a temperature increase of 10 at constant volume, what is the heat absorbed by (a) 3.0 mol of a dilute monatomic gas; (b) 0.50 mol of a dilute diatomic gas; and (c) 15 mol of a dilute polyatomic gas?A cylinder containing three moles of a monatomic ideal gas is heated at a constant pressure of 2 atm. The temperature of the gas changes from 300 K to 350 K as a result of the expansion. Find work done (a) on the gas; and (b) by the gas.Consider the Maxwell-Boltzmann distribution function plotted in Problem 28. For those parameters, determine the rms velocity and the most probable speed, as well as the values of f(v) for each of these values. Compare these values with the graph in Problem 28. 28. Plot the Maxwell-Boltzmann distribution function for a gas composed of nitrogen molecules (N2) at a temperature of 295 K. Identify the points on the curve that have a value of half the maximum value. Estimate these speeds, which represent the range of speeds most of the molecules are likely to have. The mass of a nitrogen molecule is 4.68 1026 kg. Equation 20.18 can be used to find the rms velocity given the temperature, Boltzmanns constant, and the mass of the atom or molecule. The mass of a nitrogen molecule is 4.68 1026 kg. vrms=3kBTm=3(1.381023J/K)4.681026kg=511m/s Using the results of Problem 28 and the rms velocity, we can calculate the value of f(v). f(vrms) = (3.11 108)(511)2 e(5.75106(511)2) = 0.00181 The most probable speed, for which this function has its maximum value, is given by Equation 20.20. vmp=2kBTm=2(1.381023J/K)(295K)4.681026kg=417m/s f(vmp) = (3.11108)(417)2 e(5.75106(417)2) = 0.00199 We plot these points on the speed distribution. The most probable speed is indeed at the peak of the distribution function. Since the function is not symmetric, the rms velocity is somewhat higher than the most probable speed. Figure P20.29ANS