Good question.Alex-RCHS wrote:Not to distract from the current question, but can you explain this again? If the change in the energy of the gas is not equal to the work done on the gas, then where does that "extra" work go? (For lack of better phrasing)Justin72835 wrote:Close. 1717J is actually the work done to the gas. If you want to find the change in internal energy then you can use the formula 5/2(PV). This differs from using 3/2(PV) because it is a diatomic gas, meaning that some of the energy goes into causing the molecules to vibrate rather just its kinetic energy for a monatomic gas. After finding the change in internal energy, you can actually find the heat added or removed from the gas using the First Law.Alex-RCHS wrote:1073 torr = 143055 pascals. Pascal = J/m^3 15L = 15*1000cm^3 = .015m^3 .015m^3 * 143055 J/m^3 = 2146J 3L=.003m^3 .003m^3 * 143055 J/m^3 = 429J 429J-2146J = a loss of 1717J of heat

You're asking why the work you do on the gas doesn't contribute completely to a change in internal energy. Try to look at it this way. If you actually had the piston in front of you and you tried compressing the gas by pushing down on the top of it, you would feel the pressure increase. In fact, the more you try to compress the gas, the more pressure you'll feel the gas apply back.

**This is different from the scenario**I stated before because

**the pressure is changing with the volume**.

Now imagine this scenario: You have the same piston, but you put ice around it to cool the gas. You can see the top of the piston lower naturally (see Charles' Law). What's happening in this situation is that the ice is drawing heat away from the piston (heat removed), which contributes to lowering the temperature of the gas and making its volume shrink in size. It is clear here that the pressure inside the piston remains constant, as it is still supporting the weight of the piston as well as the atmospheric pressure. This also means that the surroundings are doing work on the gas.

The second scenario is the one laid out in the problem I gave. It is clear here that the surroundings are doing work on the gas, but since it is at a constant pressure,

**the gas is actually losing heat as well**. This all results in a decrease in internal energy.

The First Law of Thermodynamics can be written as [math]\Delta U = Q + W[/math]. We already know that the work is being done on the gas since it is being compressed. The next step at this point is to find the change in internal energy of the gas. Since it is a diatomic gas, the equation [math]\Delta U = \frac {5}{2} nR \Delta T[/math] can be applied here. We know that [math]PV = nRT[/math] from the Ideal Gas Law, so we can substitute this term into the previous equation to get [math]\Delta U = \frac {5}{2} P \Delta V[/math]. After solving this out, we find that change in internal energy is -4292 J. Plugging in this value into the first equation, we can solve for the heat added/removed, which turns out to be -6008 J (heat is removed). Change in internal energy is -4291 J Heat removed is 6008 J Work done on gas is 1717 J