Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 19, Problem 33P
To determine
The PV diagram and efficiency.
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A 2.00-mol sample of helium gas initially at 300 K and0.400 atm is compressed isothermally to 1.20 atm.Assuming the behavior of helium to be that of an idealgas, find (a) the final volume of the gas, (b) the workdone by the gas, and (c) the energy transferred by heat
A monatomic ideal gas initially has a temperature of 330k and preassure of 3x105 Pa. The gas expands from a volume of 500cm3 to a volume of 1500cm3.
l) Calculate the change in internal energy (in Joules), if the expansion is adiabatic.
answer => (150)ln(3)J
NOTICE : The work in adiabatic process equals the change in internal energy. So, w = - Δu = - (g/2*n*r*Δt)
The answer above is right, and was calculated using integrals. I would like to know how to calculate using Δu = g/2*n*r*Δt, since the work in adiabatic process = -Δu(change in internal energy)
Chapter 19 Solutions
Physics for Scientists and Engineers
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- 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.arrow_forwardTwo moles of a monatomic ideal gas at (5 MPa, 5 L) is expanded isothermally until the volume is doubled (step 1). Then it is cooled isochorically until the pressure is 1 MPa (step 2). The temperature drops in this process. The gas is now compressed isothermally until its volume is back to 5 L, but its pressure is now 2 MPa (step 3). Finally, the gas is heated isochorically to return to the initial state (step 4). (a) Draw the four pi-cresses in the pV plane. (b) Find the total work done by the gas.arrow_forwardA 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)?arrow_forward
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