Is it possible to compress gas without making it hotter?

sengihantq
2022-09-01
Answered

Is it possible to compress gas without making it hotter?

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Rachael Singh

Answered 2022-09-02
Author has **4** answers

Consider an ideal gas at $300\phantom{\rule{thinmathspace}{0ex}}\text{K}$ at a pressure of 1 atm. The ideal gas law gives a density of about $2\times {10}^{25}$. So you'd expect that you'd be able to move the piston a distance of about $1/\lambda =5\times {10}^{-22}\phantom{\rule{thinmathspace}{0ex}}\text{m}$ before it collides with a gas particle. This is an incredibly tiny distance, about ${10}^{-6}$ times the size of a proton. The upshot is that it's practically impossible for the situation you're describing to occur. As soon as you move the piston at all, it will collide with a gas particle.

asked 2022-10-13

In Ashcroft and Mermin Chapter 1, just above equation (1.50) and in the context of a classical ideal electron gas, it is said that the electronic specific heat at constant volume ${c}_{v}$ is defined by

${c}_{v}=\frac{\frac{dE}{dT}}{V}$

which seemed highly irregular to me (I would have expected division by the mass M of the system of electrons). Is there any reason for this that I am perhaps missing?

${c}_{v}=\frac{\frac{dE}{dT}}{V}$

which seemed highly irregular to me (I would have expected division by the mass M of the system of electrons). Is there any reason for this that I am perhaps missing?

asked 2022-10-18

If boiling of water involves change in internal energy, then why does the temperature remain constant?

asked 2022-09-26

Why does the Gibbs free energy need to be minimized for an equilibrium?

$dG=dE+pdV-TdS$

But $dE=Tds-pdV$ so it should be $dG=0$. Why $dG$ is not always zero?

$dG=dE+pdV-TdS$

But $dE=Tds-pdV$ so it should be $dG=0$. Why $dG$ is not always zero?

asked 2022-10-15

So the heat capacity ratio is the ratio of the heat capacity at constant pressure to heat capacity at constant volume. Since there is no pressure in the vacuum, the heat capacity ratio of the vacuum should be zero. Is this right ?

asked 2022-08-06

A substance at 200-degree Celsius is given some amount of heat to raise its temperature by one degree Celsius and the same substance when at -200 degrees Celsius is given some amount of heat to increase its temperature to -199 degrees Celsius. Is the amount of heat required for both the processes same?

asked 2022-07-19

Is there a name for this sort of thermo relationship?

${H}_{vap\phantom{\rule{thickmathspace}{0ex}}/\phantom{\rule{thickmathspace}{0ex}}sub\phantom{\rule{thickmathspace}{0ex}}/\phantom{\rule{thickmathspace}{0ex}}cond}({T}^{\prime})-{H}_{vap\phantom{\rule{thickmathspace}{0ex}}/\phantom{\rule{thickmathspace}{0ex}}sub\phantom{\rule{thickmathspace}{0ex}}/\phantom{\rule{thickmathspace}{0ex}}cond}(T)={\int}_{T}^{{T}^{\prime}}\mathrm{\Delta}{C}_{p,m}\phantom{\rule{thickmathspace}{0ex}}dT$

${H}_{vap\phantom{\rule{thickmathspace}{0ex}}/\phantom{\rule{thickmathspace}{0ex}}sub\phantom{\rule{thickmathspace}{0ex}}/\phantom{\rule{thickmathspace}{0ex}}cond}({T}^{\prime})-{H}_{vap\phantom{\rule{thickmathspace}{0ex}}/\phantom{\rule{thickmathspace}{0ex}}sub\phantom{\rule{thickmathspace}{0ex}}/\phantom{\rule{thickmathspace}{0ex}}cond}(T)={\int}_{T}^{{T}^{\prime}}\mathrm{\Delta}{C}_{p,m}\phantom{\rule{thickmathspace}{0ex}}dT$

asked 2022-08-26

If we take the law of an ideal gas:

pV = nRT

We see that reducing the pressure to 0, or close to it, would lead to a similar drop in the temperature.

What would be the temperature inside of a voided (emptied of any gas) chamber? Could we use it as a cooling storage?

pV = nRT

We see that reducing the pressure to 0, or close to it, would lead to a similar drop in the temperature.

What would be the temperature inside of a voided (emptied of any gas) chamber? Could we use it as a cooling storage?