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# Calculating internal energy and work example

Worked example calculating the change in internal energy for a gas using the first law of thermodynamics.

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• Where is the 10^-3 coming from (multiplied by m^3?
• because 1L equals 10^-3 m^3. taking 1L in the denominator and 10^-3 m^3 in the numerator we aren't affecting the value but only the units.
that's how I make the meaning of this
• Why work is pressure multiplied by the change in volume?
• We get the relation using the Laws of thermodynamics,
PV = nRT = *Delta*W
where P is pressure, V is volume, n is moles of the gas, R is the gas constant, T is the temperature, W is the work
(1 vote)
• The drawing is somewhat confusing.
It appears that 485J is being moved from V1 to V2. But if I am understanding this correctly, the 485J is leaving V1 and the result is V2 correct?
• Yup. But at the same time 25.25J of work is done(added) to the v1 to become v2. So then net change of internal energy of the balloon is the heat loss(-485J) plus the work done(+25.25J).
• I'm sorry but why is it that doing work on the system increases the internal energy in the first place? What exactly does work done to the system mean? like squeezing the baloon? but that just decreases the volume and increases the pressure and the energy shouldn't change..(?) Also, if Heat, Q, was transferred to the system, can I view that as like someone heated the system with fire or something? If so, that makes sense as molecules gain kinetic energy from heat. I really appreciate if you can answer this.
• When work is done on a system, energy is transferred to that system, which increases the internal energy of the system. Conversely, energy is lost from whatever is doing the work on the system.

Heating a system with a fire is a classical way of transferring heat to the system. That was the basis of the steam engines that started the whole science of thermodynamics.
• Isn't it 1L is equal to 1dm^3?
• in simple way to remember that volume part ,just take it in 1-dimension and expand to 3-dimension like 100mm=10cm=1dm=0.1m,just cube all three we get 10^6mm^3=10^3cm^3/cc=1dm^3=10^-3m^3=1L
(1 vote)
• Why is the sign for pressure is negative"
• Go ahead and watch this part.
We are considering the work that is done on the system. So if you compress the space, you are doing work to it.
If the system (balloon) expands, it is doing work to the surrounding. In this case, the system loses energy by doing work, thus the work done is negative.
• At t=2.47, why do you add a minus sign there? It appears all of a sudden...
• That comes from the formula for "work" for a mechanically reversible system. She doesn't goes into the details, but for a mechanically reversible system:
w= -Pext(delta V). You need the negative in front because when work is done "on the system" (which is positive), delta V (Vfinal-Vinitial) is negative (the ballon is compressed). So the two negatives make the whole term for work positive. Side note: remember that external pressure (Pext) is considered constant. Which makes the formula for work so simple. So we have not learned anything in this problem about the pressure inside. That would require more assumptions.
• In the formula the negative sign indicate that the pressure of gas that is in the system is working against the external pressure . the external pressure could be block or anything and in the video the balloon compress that means work is done by the surrounding leading to increase in internal energy and the work is positive
• At about she defines joules in terms of other units(kgm2/s2) please could you tell me how she derived that?
(1 vote)
• It's one of the definitions of joules...

The energy transferred to an object when a force of 1 Newton acts on that object a distance of 1 metre.

J = N * m

N = kg * m * s^-2

Therefore: J = kg * m^2 * s^-2