What happens to the temperature when a gas is expanded adiabatically?
Gases can be expanded under adiabatic conditions from high pressure to low pressure either using-a turbine (isentropic if considering process to be reversible)a throttling device (isenthalpic process)Expansion of gases through a turbine:Steady flow expansion of a high pressure gas through a turbine or an expansion engine results in a net work output with a resulting decrease in enthalpy. This decrease in enthalpy leads to a decrease in temperature. In an ideal case, the expansion will be reversible adiabatic, however, in an actual case, the expansion can be adiabatic but irreversibility exists due to fluid friction. If the changes in potential and kinetic energy are negligible and the process is adiabatic, then:Wnet = (h1 − h 2 )Since Wnet is positive, the outlet enthalpy h2 is less than inlet enthalpy h1; hence the outlet temperature T2 will also be less than inlet temperature T1, an approximately reversible adiabatic expansion with a net work output always produces a decrease in temperature irrespective of the initial temperature.Expansion of gases by throttling: The throttling of gases is an isenthalpic process.Since the enthalpy of an ideal gas is a function of temperature only, during an isenthalpic process, the temperature of the ideal gas remains constant.In case of real gases, whether the temperature decreases or increases during the isenthalpic throttling process depends on a property of the gas called Joule-Thomson coefficient, μJT, given by:Now see the different regions in the p-T diagram of the gases-Along a constant enthalpy line (isenthalpic process), beginning with an initial state ‘i’ the temperature of the gas increases initially with reduction in pressure upto point f3, and μJT is negative from point i to point f3. However, further reduction in pressure from point f3 to f5, results in a reduction of temperature from f3 to f5. Thus point f3 represents a point of inflexion, where μJT = 0. The temperature at the point of inflexion is known as inversion temperature for the given enthalpy. Therefore, if the initial condition falls on the left of inversion temperature, the gas undergoes a reduction in temperature during expansion and if the initial condition falls on the right side of inversion point, then temperature increases during expansion.Footnotes:http://nptel.ac.in/courses/Webco...
What would happen if you left a gas (or electric) stove unattended for several hours?
This isn't theoretical, it happens not uncommonly, both accidentally and on purpose.Both gas and electric stoves automatically control their output and will run at a more or less constant temperature indefinitely. That means you'll waste some gas or electricity, and mildly heat the house.My mom once accidentally left a pot of water boiling on the stove over a low heat and went to bed. When she woke up, all the water had boiled away, and the pot had been warped into uselessness by being heated while dry for several hours. On another occasion, I left my apartment having forgotten I was boiling beans on the stove. When the water boiled away, the beans burned and fused into the pot, and my landlord went in because the fire alarm was going off. The place smelt terrible and the pot was a total loss, but nothing else happened.In short, I wouldn't recommend doing it, but nothing happens except that the stove stays hot, and so does whatever's on it.
An open flask of air is heated, stoppered in the heated condition, and then allowed to cool back to room temp,
The flask contains fewer gas molecules now than it had before it was heated. The reason is that heating the flask causes the gas molecules inside it to heat up, and as they heat up, each gas molecule takes up more volume. Pressure increases, but without a stopper, gas simply leaks out of the flask - a decrease in the amount of gas molecules present. When the stopper is placed over the opening and the flask removed from the heat, everything starts to cool down. As the gas molecules cool, pressure decreases, but there is no way for gas to re-enter the flask. Thus the flask is trapped with less gas molecules than before. The volume occupied by the gas in each instance is the same - the volume of the flask does not change. The pressure in the flask is lower after it was stoppered and cooled, as mentioned before. Finally, do any of the above conditions explain why water rushed into the flas at the lower temp? Well, you didn't mention the water, or the events and conditions surrounding its possible entry into the flask, so...all I can say is that reduced pressure inside the flask would create a vacuum there, and if water were to be introduced to this system then it certainly would get sucked into the flask
If air is heated in a sealed container what will be change in density?
None. You see density = [math]mass/volume[/math] You must know that mass cannot change as the container is sealed, so no molecules can leak out. You must also that volume of the container is constant. So, when mass and volume are same as they were before heating, then does density changes after heating? No!!BUT, I think what made you confuse here was the thought that air becomes less dense on heating, but that is valid of open spaces, because the air can expand to the extent it wants to.The things changing here is the kinetic energy of the molecules of air and their temperature, the pressure inside the container. :)
Calculate the work done (in joules) by the gas if it expands? Chem help?
W = PdV = 0atm x (0.939L - 0.263L) = ___ Latm and W = PdV = 4.00atm x (0.939L - 0.263L) = ___ Latm then.. if you recall the two common values of R... R = 8.314 J/moleK R = 0.08206 Latm / moleK so.. 8.314 J = 0.08206 Latm.. and to finish the problem.. __ Latm x (8.314 J / 0.08206 Latm) = __ J ******* questions?
670 J of work are done on a system in a process that decreases the system's thermal energy by 10 J. How much?
Change of system's internal energy equals work done On it plus heat heat transferred to it. ∆U = W + Q thus, Q = ∆U - W = -10J - 670J = -680J So 680J of heat are transferred. Negative sign indicates that heat is transferred from the system.