If we compress the air in a cylinder from 1/2 atm to 1atm, would the air inside get hotter (than what it was before compressing)? If so, by how many degrees? (Considering an ideal piston and cylinder that do not produce or absorb heat) If we try the same experiment with freon instead, how would it compare to the air? (In terms of temperature change) Both questions are of interest to me, but I really need to know how many degrees the temp of the air in the first experiment will rise. Thank you
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$\begingroup$ Yeah the air would get hotter. You can use the gas equation to find out by how much.( remember that in the gas equation pressure is directly proportional to temperature). According to google the density of Freon is more than air. So expect that the temperature of air would be more than that of Freon if the experiment is conducted in the same manner. $\endgroup$– physics2000Commented Nov 4, 2018 at 16:06
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$\begingroup$ Thank you Physics2000. Why is air not used in heating and cooling systems instead of freon ? $\endgroup$– Alex DoeCommented Nov 4, 2018 at 16:51
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$\begingroup$ well here’s the thing about Freon. First and foremost it’s non toxic( true for humans but not for the ozone layer). Second it’s easy to condense. Just imagine trying to condense air. Air is a mixture of Nitrogen, CO2 , water and others. It’s not impossible just that it’s very inefficient to do it. $\endgroup$– physics2000Commented Nov 4, 2018 at 17:03
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$\begingroup$ A refrigeration system necessarily requires that you can condense the refrigerant at a temperature that is only 20 deg F or so higher than the heat sink temperature. For air conditioning, this is approximately 120 deg F. It is impossible to condense air at this temperature, as 120 deg F is far above the critical temperature of air, so air cannot be used in a traditional refrigeration system. $\endgroup$– David WhiteCommented Nov 4, 2018 at 17:54
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For an adiabatic change in pressure (adiabatic means no heat added or removed from the gas) $P^{1-\gamma}T^\gamma$ remains constant during the process, where $P$ is pressure, $T$ is temperature (in Kelvin) and $\gamma$ is the ratio of specific heats ($\frac{7}{5}$ for air). This means, if you increase the pressure of a parcel of air by a factor of 2 (from half an atmosphere to 1 atmosphere), you change the temperature by a factor of $2^{2/7}$, or $1.219$. So, if the air at half an atmosphere has a temperature of freezing (273 K), then after pressurization to one atmosphere, the temperature will be about 333 K. That's 60 C or 140 F: quite hot.
I don’t know $\gamma$ for freon, but it’s sure to be smaller than for air because it’s a 3-dimensional molecule. So freon would not increase in temperature quite as much.
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$\begingroup$ Ben51, in my opinion, an adiabatic argument for freon compression doesn't make a whole lot of sense. Refrigeration systems require heat rejection to the environment, so the OP's question regarding temperature rise due to the heat of compression isn't particularly relevant to systems that use freon. $\endgroup$ Commented Nov 5, 2018 at 3:02
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$\begingroup$ @DavidWhite I don’t understand what you mean. I don’t know anything about freon, but if there’s a temperature at which it can be compressed from half to one atmosphere without condensing I don’t see why it shouldn’t obey the gas laws. $\endgroup$– Ben51Commented Nov 5, 2018 at 11:15
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$\begingroup$ First, the ideal gas law applies to ideal gases, and freon is not an ideal gas. Second, freon is used in refrigeration systems. Such systems are designed with a condenser area that is sufficient to transfer the heat of compression to the environment AND condense the freon. Nobody who designs such a system would ever give any consideration to adiabatic conditions for freon, because you can't compress freon AND condense it under adiabatic conditions. $\endgroup$ Commented Nov 5, 2018 at 16:04
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$\begingroup$ @DavidWhite Does not seem relevant how freon is used. No mention of that in OP’s question. And departures of real gasses from ideal behavior near room temperature and pressure are slight. $\endgroup$– Ben51Commented Nov 6, 2018 at 14:46