# Why is heat measured in joules?

Andrew Mason
Homework Helper
I don't get how is the isothermal process done ???
I
If T is constant and V increases then P has to decrease. If P decreases, the walls do not expand if the external pressure remains the same. So if T is constant, the external pressure has to decrease if there is an expansion.

If you take your example, if heat flows into a gas, T increases and P increases if V is constant. However, if the external pressure is constant, then an increase in V continues until P is equal to the external pressure. So this describes a constant pressure expansion.

AM

If T is constant and V increases then P has to decrease. If P decreases, the walls do not expand if the external pressure remains the same. So if T is constant, the external pressure has to decrease if there is an expansion.

If you take your example, if heat flows into a gas, T increases and P increases if V is constant. However, if the external pressure is constant, then an increase in V continues until P is equal to the external pressure. So this describes a constant pressure expansion.

AM
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I think I'll have to study the gas laws first, let's start with Boyle's law

It states that pressure is inversely proportional to volume, is that because the no of collisions of the molecules with the walls of the container due to Brownian motion increase in a smaller volume ???

jbriggs444
Homework Helper
2019 Award
It states that pressure is inversely proportional to volume, is that because the no of collisions of the molecules with the walls of the container due to Brownian motion increase in a smaller volume ???
It's not that the motion of each molecule increases. If temperature is held contstant, the average velocity of the molecules will not change. It is that there are more molecules. So more molecules hitting a given area of wall surface per unit time.

It's not that the motion of each molecule increases. If temperature is held contstant, the average velocity of the molecules will not change. It is that there are more molecules. So more molecules hitting a given area of wall surface per unit time.
Yeah I got that, It just increases the chance of its collision with the walls, we can say it changes the no of collisions per unit area and per unit time but the average kinetic energy doesn't change since the Temperature is constant

when we are talking about the pressure of gases, what is really meant, its pressure due to gravity?? or its pressure due to its K.E and its collision with the surroundings ????

Compress (or allow to expand) a gas in a situation where the temperature cannot change and you get Boyle's Law behaviour. PV is constant.

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If I compressed a gas in a container of movable walls which isn't thermally isolated from the surrounding, that is considered as a isothermal process ??

sophiecentaur
Gold Member
If I compressed a gas in a container of movable walls which isn't thermally isolated from the surrounding, that is considered as a isothermal process ??
It would be far more efficient for you to read through and follow what a text book has to say about all this. Look at This link and also Wikipedia will tell you all you need to know. Q and A is a very poor way of learning basic stuff, compared with going with the flow of the argument presented in a text book - that has been tried and tested. Q and A is a bit like Brownian Motion; every so often you may randomly go in the right direction but most of the time you can end up off-target.

stevendaryl
Staff Emeritus
If I compressed a gas in a container of movable walls which isn't thermally isolated from the surrounding, that is considered as a isothermal process ??
Isothermal means that it is kept at a constant temperature. One way to do this is to have a very large reservoir (or air, or water, or whatever) that is kept at a constant temperature, and do the compression so slowly that the gas in the container is always allowed to reach equilibrium with the reservoir.

Isothermal means that it is kept at a constant temperature. One way to do this is to have a very large reservoir (or air, or water, or whatever) that is kept at a constant temperature, and do the compression so slowly that the gas in the container is always allowed to reach equilibrium with the reservoir.

So during the slow compress the gas keeps gaining and losing heat to the surrounding medium at the same rate staying at an equilibrium position with surrounding, so its temperature stays constant

stevendaryl
Staff Emeritus
So during the slow compress the gas keeps gaining and losing heat to the surrounding medium at the same rate staying at an equilibrium position with surrounding, so its temperature stays constant
Normally, compressing a gas would cause it to get hotter. But you can view isothermal compression as a limiting case of the following process:

Compress the gas a tiny bit.
Let it cool off.
Compress it a tiny bit more.
Let it cool off.
Etc.

The resulting pressure when you reach your final volume will be less than it would have been if you hadn't allowed it to cool off.

Isothermal compression results in this relationship between pressure and volume:

$P \propto V^{-1}$

If the gas were prevented from exchanging heat with the reservoir, the relationship would be:

$P \propto V^{-5/3}$ (for a monatomic gas).

sophiecentaur
Gold Member
So during the slow compress the gas keeps gaining and losing heat to the surrounding medium at the same rate staying at an equilibrium position with surrounding, so its temperature stays constant
If you change the volume 'slowly enough' the temperature can remain 'near enough' constant. It's an ideal scenario. The other extreme case - adiabatic volume change, assumes that no heat enters of leaves but that again is only an ideal situation. Real life is always somewhere in between.

If you change the volume 'slowly enough' the temperature can remain 'near enough' constant. .

If it's not done slowly it will gain heat, but eventually it will lose it reaching to an equilibrium state with the surrounding medium, so the final result is the same whether it's done slowly or fast, why should be so careful about that point ??

WannabeNewton
The microscopic kinetic energy of the molecules does not correspond to temperature. The quantity that corresponds to temperature is ##\frac{\partial{E}}{\partial{S}}## where ##S## is the entropy...
For a classical gas the equipartition theorem states that the kinetic energy of each mole of the gas is directly proportional to the temperature of the gas. The definition ##T = \frac{\partial \bar{E}}{\partial S}##, or even better ##S = -\frac{\partial F}{\partial T}## in the Helmholtz representation, does not contradict this as the equipartition theorem is proven using the statistical equivalent of that latter definition i.e. ##\bar{E} = -\frac{\partial}{\partial \beta}\ln Z##. Sure the kinetic energy is itself not temperature but for classical gases it is directly proportional to it as are the energies of other quadratic degrees of freedom.

To the OP: heat ##\delta Q## measures the extent to which the change ##\Delta U## in the internal energy of a system through a process fails to be adiathermal. As others have noted, one can perform simple experiments to show that a quantity ##\delta Q## must exist to account for the change in internal energy of a system between equilibrium states that is not already taken into account by the work done on the system during the process and we know such a quantity must exist simply due to the existence of processes which are the entire opposite of adiabatic and which involve no work done at all but still have a change in internal energy (just take a beaker of water and put a lit Bunsen-burner under it). Clearly then ##\delta Q## must be measured in Joules.

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stevendaryl
Staff Emeritus
If it's not done slowly it will gain heat, but eventually it will lose it reaching to an equilibrium state with the surrounding medium, so the final result is the same whether it's done slowly or fast, why should be so careful about that point ??
Well, if the temperature is (approximately) constant, then the process is (approximately) isothermal. It's just a matter of definition.

You're right that for an ideal gas, the final state doesn't depend on whether the process was isothermal, or not. But the total work done and the change in the entropy of whatever system was compressing the gas depends on whether it was isothermal or not.

During his honeymoon in Switzerland Joule along with Thompson, (history does not relate why Joules and Thompson happened to meet in Switzerland during Joule's honeymoon), attempted to measure the temperature difference between the top and bottom of a waterfall, history also fails to mention what Joule's bride Amellia thought of it.

stevendaryl
Staff Emeritus
During his honeymoon in Switzerland Joule along with Thompson, (history does not relate why Joules and Thompson happened to meet in Switzerland during Joule's honeymoon), attempted to measure the temperature difference between the top and bottom of a waterfall, history also fails to mention what Joule's bride Amellia thought of it.
That's funny. Maybe the story was twisted for the sake of New Yorkers like me, but I thought t was Niagara Falls.

Well, if the temperature is (approximately) constant, then the process is (approximately) isothermal. It's just a matter of definition.

You're right that for an ideal gas, the final state doesn't depend on whether the process was isothermal, or not. But the total work done and the change in the entropy of whatever system was compressing the gas depends on whether it was isothermal or not.

In case of the isothermal process, If the walls of the container are movable, and then the gas is compressed, part of the energy will be used up to expand the volume and the other part will increase the gas's temperature which will eventually get lost to the surrounding reaching a state of thermal equilibrium ( the temperature won't change ) ???

stevendaryl
Staff Emeritus
In case of the isothermal process, If the walls of the container are movable, and then the gas is compressed, part of the energy will be used up to expand the volume and the other part will increase the gas's temperature which will eventually get lost to the surrounding reaching a state of thermal equilibrium ( the temperature won't change ) ???
If it's isothermal, then none of the energy goes into raising the temperature of the gas.

Some of the energy required to compress it will go towards decreasing the volume, and some of the energy will go towards heating up the reservoir (by assumption, the reservoir is so large that this extra heat causes no significant change to the temperature of the reservoir.

sophiecentaur