(II) An ideal gas expands at a constant total pressure of 2.5 ...

Question

(II) An ideal gas expands at a constant total pressure of 2.5 atm from 410 mL to 690 mL. Heat then flows out of the gas at constant volume, and the pressure and temperature are allowed to drop until the temperature reaches its original value. Calculate
(a) the total work done by the gas in the process, and
(b) the total heat flow into the gas.

(a) the total work done by the gas in the process, and

(b) the total heat flow into the gas.

Accepted Answer

Hi everyone. Let's take a look at this practice problem dealing with work and heat transfer. So in this problem, a sample of helium gas expands under constant pressure conditions of about four atmospheres from an initial volume of 500 mL to a final volume of 950 mL. The gas then cools down at constant volume until it returns to its original temperature state. Again, we need to calculate how much work was performed during these processes and find out how much heat was transferred. We're given four possible choices as our answers. For choice. A work is equal to negative 590 joules and Q is equal to negative 590 joules for B work is equal to negative 590 joules and Q is equal to 590 joules. For choice C work is equal to 100 80 joules and Q is equal to negative 180 joules. And for choice D work is equal to 100 80 joules and Q is equal to 100 80 joules. So the first part of this question wants us to calculate the work total work performed by these processes. So that's where we'll start. So our total work W is going to be equal to the work done. Under each step, we'll have W one plus W-2, W one and W-2 are the works done by the 1st and 2nd processes respectively. So in order to calculate this, we need to recall our formula for the work done by a gas and recall that that formula is W is equal to P multiplied by DV. We here P is pressure and DV is our change in volume. So we're gonna use this formula to calculate the W one and W-2. So we'll start with the W one. So W one is going to be equal to, for a pressure. Here, our initial pressure is the four atmospheres and it's constant have four atmospheres. But I'll need this in pascals. So I'll multiply this by the conversion factor of 1.01 multiplied by 10 to the five pascals. And this gets divided by one atmosphere. And this pressure here gets multiplied by our change in volume. So I'll have my final volume, which is the 950 mL. And I'll subtract from it our initial volume, which is the 500 mL. However, I'm gonna need to convert this into meters cubed. So I'll multiply this by conversion factor of 1 m cubed divided by one multiplied by 10 to the 6 mL. So I can plug these values into my calculator. And again, W one is equal to 181.8 joules. Now we do the same thing for W-2. So W-2 is gonna be equal to our pressure p multiplied by our change in volume. But the second process is at constant volume. So my change in volume is gonna be zero. So that means that my W-2 is equal to zero joules. So don't need to worry about figuring out what the pressure is and how that pressure changes with the temperature. Since it's always gonna be multiplied by zero, I just get zero joules. So now I can combine these two together to find our total work will have W is equal to, for W one, we'll have the 181.8 joules plus zero joules or W-2. And here, since my answer choices only have two significant figures in them, I'm going round by total work to two significant figures. So that means W is going to round 2, 100 80 jewels. So that's the first part to our answer. Now, for the second part, we need to calculate how much heat was transferred for this. We'll want to use our first law of thermodynamics to recall that for the first law of thermodynamics, we have our change in internal energy delta E internal and that is equal to our heat Q minus the work W. So we're gonna solve this for Q. So I have Q is equal to delta E internal plus W. So we've already calculated the work, but I now need to calculate my change in internal energy. And once I do that, um I will be able to find our heat. So recall your definition for the change in internal energy, the data E internal is equal to N CV. Delta T we're in, here's the number of moles CV is the molar specific heat at constant volume and delta T is our change in temperature. Now, we were told the problem that we end at the same temperature that we begin. So that means that my delta T here, it's gonna be equal to zero. And since my change in temperature is zero, that means my change in internal energy delta E internal is gonna be equal to zero joules. So I can plug that value in along with our work value to calculate our heat. And our first law of thermodynamics formula coming back to that formula, we'll have Q is equal to zero joules plus the 180 jewels. And so that means that Q is equal to 180 jewels. And that is the second part to my answer. So now I calculated the work and the heat I can compare those to the answer choices that I was given. And the problem and this corresponds to answer choice D. And here we want to point out that the work was positive. So that means that the work was being done by the gas. If we'd had a negative work. That means that the work would be have been done on the gas. Likewise, with the um heat being positive, that means that that heat is flowing into the gas. If I had calculated a negative heat, that means that the heat would have been flowing out of the gas. So just a quick little recap of what we did. We used our definition of work due to a gas to calculate the total work done by each of these uh two processes. And then we used our first law of thermodynamics to calculate the heat that was transferred during the um two processes. But in order to do that, we had to calculate first or change the internal energy using that definition. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Ideal Gas Law

Work Done by a Gas

First Law of Thermodynamics

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