Here's the whole paragraph for the isothermal expansion ... explain. thanks.?

Question

In certain cases, we can relate the value of q to the change in volume of a system and so can calculate, for instance, the flow of energy as heat into the system when a gas expands. The simplest cas is that of a perfect gas undergoing isothermal expansion. Because the expansion is isothermal, the temperature of the gas is the same at the end of the expansion as it was initially. Therefore, the mean speed of the molecules of the gas is the same before and after the expansion. That implies in turn that the total kinetic E of the molecules is the same. But for a perfect gas, the only contribution to the energy is the kinetic E of the molecules; so, we have to conclude that the total E of the gas is the same before and after the expansion. E has left the system as work; therefore, a compensating amount of E must have entered the system as heat. We can therefore write "For the isothermal expansion of a perfect gas: q = -w "

Here is my argument/understanding: Pressure is the capacity to do work. (Pressure is Force) PV=nRT. Volume increase, however, (the result of the P already having done work) is not work; it is maximum entropy. (the Force (as Pressure) has done work on the surroundings. Therefore, deltaV is a method of measuring work performed. This work performed is ultimately a function of the increase in heat. However, although we assume deltaP=0, the entropy increase (increase in V) of the gas is resultant of increase in T.

THEY TOLD YOU THAT THEY HEATED THE GAS!!!! So, in PV=nRT, the T increases. On the left side of the equation, either the P or the V MUST increase!

If there is an empty coke bottle in my refigerator and there is a thermometer in the bottle, and a "cylinder" screwed onto the cap of the bottle, and THEN i remove it from the fridge, and the cylinder expands (V increases), the thermometer temp will rise!!! In other words, V cannot expand unless there is an accompanying increase in T!!!

If x moles (constant) of gas contains a larger V, then it has to have been heated! True!

Answer 1

First, forget about the entropy change. It's there, but it has nothing to do with what they're telling you here. You're on the right track about pressure and work. For an isothermal system, the change in PV is equal to the work done.

Now the normal expectation for an expanding gas is that it cools off. But they have told you that heat has entered the system, so the process is not adiabatic or reversible. Finally, they conclude that the amount of heat put into the system must be equal to the amount of work taken out of the system. The reasoning is that the the kinetic energy of the gas is unchanged, because its temperature is unchanged (isothermal), and the only energy transactions are heat in and work out. Therefore the two must be of equal magnitude.

What they haven't told you is that the entropy and enthalpy have changed, but I'm guessing they'll get to that soon enough.

Answer 2

Okay, settle down. ISOTHERMAL = constant temperature T does not change. Period.

The introduction of head ENERGY into the system causes the pressure to change. The change in pressure causes the work, resulting in the expansion of the volume. Remember a force on a mass at rest is not performing work.

Answer 3

The whole idea with isothermal expansion is that the temperature does not change. Isothermal means one temperature.

I think your problem is with the word heating. You assume that because you are heating something is temperature is going to increase. This is not always the case. Think of boiling water, as long as the water is boiling you can continue to add heat to it but its temperature will not change. It will remain at 100 C.
In this sense heating is just adding Energy to the system. This is what happens in isothermal expansion.

Answer 4

sorry if impolite but we EXPLAINED it is the UNDERSTANDING that is missing BUT NOTE the term PERFECT GAS it is equivalent to INFINITY a concept not a reality this is a FORMULA not a description but the answer remains the same THE HEAT IS SHED AS FAST AS IT IS PRODUCED SO NO GAIN FROM THE EXPANSION OF THE GAS----- IN THE TEMP OF THE GAS ( HEAT WAS TRANSFERRED INTO WORK !!!!!!!!!!!!!!!!!)-----------------you are limiting your equation to the gas and not considering the surroundings there is a heat gain ( as work ) JUST NOT IN THE GAS you could use the same equation to prove that ice was impossible because the heat would always be there !!!!! anway wish you luck with this take heart tho you don't have to understand a formula to use it ( or we would all have to be Einstein do you UNDERSTAND E=MC2 ?

Answer 5

so... what's the question?