Internal energy of gas homework

[Kawrae]

>> An ideal gas is compressed to one-fourth of its original volume while its temperature is held constant. (Hint: solve part (b) first.)
(a) If 950 J of energy is removed from the gas during the compression, how much work is done on the gas?
(b) What is the change in the internal energy of the gas during the compression? <<

I'm really not even sure where to begin here at all. I know the formula for work is W=Qh-Qc... but I am not sure how to find the change in internal energy?

 

[vitaly]

Try using this link. It sends you to a problem on a college physics site; your question is very similar to the example found on the website. I hope it helps.

http://zebu.uoregon.edu/~probs/therm/scuba/scuba.html

 

[wais]

To solve this problem, we need to use the First Law of Thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat (Q) added to the system minus the work (W) done by the system. Mathematically, it can be written as ΔU = Q - W.

In this case, we are told that the temperature of the gas is held constant, which means there is no change in internal energy due to temperature. This also means that the heat added to the gas (Q) is equal to the work done on the gas (W).

(a) To find the work done on the gas, we can use the formula W = -PΔV, where P is the pressure and ΔV is the change in volume. Since the gas is compressed to one-fourth of its original volume, ΔV = -3V, where V is the original volume. We can also assume that the pressure remains constant, since the problem does not mention any change in pressure. Therefore, we can write W = -PΔV = -P(-3V) = 3PV.

Now, to find the value of P, we need to use the Ideal Gas Law, which states that PV = nRT, where n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. Since the temperature is held constant, we can write PV = nRT = constant. This means that the initial pressure (P1) and volume (V1) are related to the final pressure (P2) and volume (V2) by the equation P1V1 = P2V2. Since the gas is compressed to one-fourth of its original volume, we can write P1V1 = P2(1/4)V1. From this, we can solve for P2: P2 = 4P1. Substituting this value into our equation for work, we get W = 3PV = 3(4P1)V = 12P1V.

Now, we are given that 950 J of energy is removed from the gas during the compression. This means that Q = -950 J, since heat is being removed from the system. Since Q = W, we can write -950 J = 12P1V. Solving for P1V, we get P1V =