An experimenter adds 970 J of heat to 1.75 mol of an ideal gas to...

Question

An experimenter adds 970 J of heat to 1.75 mol of an ideal gas to heat it from 10.0°C to 25.0°C at constant pressure. The gas does +223 J of work during the expansion. (b) Calculate γ for the gas.

Accepted Answer

Hey everyone in this problem. We have a heat engine that contains two moles of an ideal gas can find in a cylinder with a movable piston. A fuel combustion provides 873 jewels of heat that increases the temperature of the gas from 45° to 60°C under constant pressure. The heated gas expands and does 249 jewels of work on the piston. Were asked to calculate the ratio of the heat capacities Gamma for the gas. Okay. All right, so let's start with what we're trying to find. Okay, we're trying to find the ratio of heat capacities, Gamma. And let's recall that this is going to be equal to the heat capacity under constant pressure. C. P divided by the heat capacity are constant volume CV. Okay, so what we need to do is we need to find cp we need to find CV in order to find gamma. Alright, let's write out what were given in the question. Okay, we know that N. Is equal to two more. Given that we have two moles of the gas. Okay, we're told that we have 873 jewels of heat. Okay, so what that means is that Q is going to be equal to jewels. We're told that the temperature changes from 45°C to 60°C. So T1 is equal to 45°C and if we add 273.15, we convert this to Calvin. We get 318.15 Calvin and same for T. Two T. Two is equal to 60 degrees Celsius, add 273.15 to get 333.15 kelvin. The heated gas expands and has 249 jewels of work. Okay, so the work W. is equal to 249 jewels. All right, well how can we find cp or cv? Let's recall that we have delta U. Equals and C. V. Delta T. Okay, so we know N. We know delta T. And we want to find C. V. How can we find delta U. Let's recall that. We also have the relationship DELTA U. Is equal to the heat Q minus the work W. Okay, well we know Q. And we know W. So we can find the change in internal energy DELTA U. It's going to be equal to 873 jewels. one is 249 jewels Which gives Delta U. equal to 624 jewels. Alright, let's go ahead and use that in the equation on the left hand side. So now we know delta U. The change in internal energy 624 jewels. And the number of moles is too Okay, two more times C. V. The value we're looking for times delta T. Which is going to be T. Two minus T. One. So we get 333.15 Calvin -318.15 Calvin. Okay, just do a little line here to separate the equations. All right, 333.15 minus 318.15. That's gonna give us 15 Calvin. Okay, so we have two moles times 15 Calvin. So we're gonna have C. V. Is equal to 624 jewels divided by 30 in our unit mole times Calvin. Okay, so more Calvin. And this is gonna give us a CV value of 20. and our unit is jewel per mole Calvin. Okay, alright, so this is the value of C. V. Now how can we find cp? Well, C p. Remember this is an ideal gas. Okay, so because this is an ideal gas, recall the way of the following relationship C. P is equal to Cv plus our Well we know Cv. We just found it 20.8 joules per mole kelvin and ours. Our gas constant 8.314 same unit, jewel per mole kelvin. If we give ourselves a little bit more room to work, This means that we have c. p value of 29. jewels per mole kelvin. So we found C. V. We found cp. Now we can go ahead and find the ratio of the heat capacities gamma by dividing the two. So gamma is equal to cp over cV Which is equal to 29.114 joules per mole kelvin Divided by 20.8 Jules per more Calvin. Okay. The units will cancel left with a unit less quantity 1.4. Okay, so the ratio of heat capacities that we were looking for is 1.4. We go back up to our answer choices. We see that we have answer D gamma is equal to 1.4. That's it for this one. Thanks everyone for watching. See you in the next video.

Struggling with Physics?

An experimenter adds 970 J of heat to 1.75 mol of an ideal gas to heat it from 10.0°C to 25.0°C at constant pressure. The gas does +223 J of work during the expansion. (b) Calculate γ for the gas.

Watch next

First Law of Thermodynamics practice set