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Ch.9 - Thermochemistry: Chemical Energy

Chapter 9, Problem 56

When a sample of a hydrocarbon fuel is ignited and burned in oxygen, the internal energy decreases by 7.20 kJ. If 5670 J of heat were transferred to the surroundings, what is the sign and magnitude of work? If the reaction took place in an environ- ment with a pressure of 1 atm, what was the volume change?

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hi everyone. This problem reads the combustion of a gas sample with excess oxygen resulted in a 10.3 kg Joel loss and internal energy and the transmission of 6480 jewels of heat to the surroundings. A identify the magnet And sign of the work done be if the external pressure is .955 atmospheres, calculate the change in volume in leaders. So we have a two part question here. The first part we want to identify the magnitude and sign of the work done and then we want to calculate the change in volume in leaders. So in order for us to calculate the magnitude of the work done magnitude and sign of the work done, we need an equation that relates change in energy heat and work. And that equation is changing energy is equal to heat plus work. Alright. And at constant external pressure heat is going to equal change in entropy. Alright, so this is at constant external pressure. Okay. So if we if we replace heat with entropy change, what we'll get when we rewrite our equation is change and energy is equal to entropy change plus work. Alright? So for our heat here we need to pay attention to the sign if heat is positive, that means the system gains takes in or absorbs heat or energy. If he is negative, that means the system loses evolves gives off or releases heat or energy. So let's go ahead and write out what's given in this problem. So what's given is we're told that the heat is equal to 6480 jewels. And we're also told that the change in energy is equal to negative 10.3 killer jewels. And this is negative because we have a loss in internal energy. Okay so another thing we can make note of is work is equal to negative pressure times change and volume. So if we replace work with negative pressure times change in volume. What our equation now comes out to is changing energy is equal to entropy change minus pressure times volume. So we replaced heat with entropy change and we replaced work with negative pressure times volume. So this is going to be our new equation here. Alright. And what we're going to do now is isolate this equation so that we're solving for entropy change. Alright. So when we isolate this equation so that we're solving for entropy change. What we get is entropy change is equal to change and energy plus pressure times change in volume. So what we did was we took this pressure times change in volume and brought it to the other side making it positive. So now what we can do is we can plug in the values that we're given. Okay because we know that change in entropy is also equal to heat. The heat was given in the problem which we wrote up above is 6480 jewels. Okay so let's go ahead and convert our heat to kill a jules because our internal are changing energy is in kilo jewels. So we're gonna want the same units. Okay, so let's go ahead and convert our heat from jewels to kill a jewels in one. Kill a jewel there is 1000 jewels. So our units for jewels cancel and we're left with killer jewels. So we're going to take 6480 and divide it by 1000. And when we do that we're going to get 6. killer jewels. And now our heat and our energy change is in the same units. So let's go ahead and plug in our values. So we're going to get we're going to get or heat it is -6.48 kg jewels. And this is negative because the problem says its heat to the surroundings. Okay, so that value is negative And then our internal our energy change is negative 10.3 kg jewels plus pressure, our pressure times volume. We don't know. So we're going to solve for this. Okay, so when we isolate our pressure times volume, let's go ahead and bring this to one side. So pressure times volume is going to equal negative 6.48 kg jewels and we're going to bring this to this side. So this is going to become positive 10.3 kg jewels. All right, so our pressure times volume is equal to 3.82 kg jewels. And remember we said appear that work is equal to negative pressure times volume. So this value that we just calculated is the work. Okay. And so our answer for the first part. So we're going to write down work is equal to negative pressure times volume. So because we need to put a negative in front of it, this becomes negative 3.82. Kill it jules. So that is the answer to the first part of the problem. Identify the magnitude and sign of the work done. So the magnitude is 3.82 kg jewels and the sign is negative. So our work done is negative 3.82 kg jewels. All right. So the second part of the question asks us to, well it says if the external pressure is 0.955 atmospheres calculate the change in volume in leaders. All right, So we know that. So now we're going to do part two. We know that work is equal to pressure times change in volume. So now we want to calculate change and volume. So to isolate that we're going to divide both sides by pressure. Alright. And so what that means is change and volume becomes work divided by pressure. We know what the work is because we just calculated that. Okay. And the work is equal to The work that we calculated is 3.8 to kill a jules. Okay, but we can't use the unit of Kila Jules. We need to convert this killer jewels to leaders times atmosphere. Alright because this is why when we plug in The pressure we know what the pressure is. The pressure was given in the problem and the pressure is 0.9 5 5 atmospheres. Okay, so we need the units for the change in volume to be in Leaders. And the way that we're going to get this atmosphere to cancel is if the units of work is in Leaders Times Atmosphere by putting the units and Leaders Times Atmosphere, our atmospheres will cancel and we'll be left with Leaders. So that means we need to convert our work to Leaders Times Atmospheres from killer jules. Okay so let's go ahead and do that. We have 3.8 to kill a jewels and we need to go from killer jewels to Leaders Times Atmosphere. So first let's go from killer jewels. Two jewels. Okay, so in one kill a jewel there is jewels. All right, so our units for kilo jewels canceled. And then the problem we were given the conversion of how to get. No we weren't given it. Okay, so the unit that we're going to use and this is something we should know is one Leader atmosphere Is equal to 101.325 jewels. If you don't know this conversion this is a conversion to make a note of. Okay, so as you can see our unit of jewels cancel. And now our unit left is leaders atmosphere. So we're going to take 3.82 and multiply it by 1000 And then divide by 101.325. And so what we get as the answer is 37.7 L atmosphere. So this is the value we're going to plug in here. All right. So we're going to have 37. Or let me write it. Okay, so we have 37.7 L atmosphere and as you can see our atmospheres cancel and we're left with leaders. So now all we have to do is do the calculation 37.7 divided by .955 gives us 39. L. Okay, so the change in volume is equal to 39.5 liters. And this is going to be our final answer for part B. Okay, so part A. The magnitude and sign of the work done is negative 3.8 to kill a jules and part B. The change in volume And leaders if the external pressure is .955 atmospheres is 39.5 L. That is it for this problem. I hope this was helpful
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Textbook Question

A reaction inside a cylindrical container with a movable RAN piston causes the volume to change from 12.0 L to 18.0 L while the pressure outside the container remains constant at 0.975 atm. (The volume of a cylinder is V = pr2h, where h is the height; 1 L # atm = 101.325 J.) (b) The diameter of the piston is 17.0 cm. How far does the piston move?

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Textbook Question

At a constant pressure of 0.905 atm, a chemical reaction takes place in a cylindrical container with a movable piston having a diameter of 40.0 cm. During the reaction, the height of the piston drops by 65.0 cm. (The volume of a cylinder is V=pr2h,wherehistheheight;1Latm=101.3J.) (a) What is the change in volume in liters during the reaction?

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Textbook Question

At a constant pressure of 0.905 atm, a chemical reaction takes place in a cylindrical container with a movable piston having a diameter of 40.0 cm. During the reaction, the height of the piston drops by 65.0 cm. (The volume of a cylinder is V=pr2h,wherehistheheight;1Latm=101.3J.) (b) What is the value in joules of the work w done during the reaction?

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What is the difference between the internal energy change ∆E and the enthalpy change ∆H? Which of the two is mea- sured at constant pressure and which at constant volume?
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