Skip to main content
Ch 19: The First Law of Thermodynamics

Chapter 19, Problem 19

Propane gas (C3H8) behaves like an ideal gas with g = 1.127. Determine the molar heat capacity at constant volume and the molar heat capacity at constant pressure

Verified Solution
Video duration:
4m
This video solution was recommended by our tutors as helpful for the problem above.
1141
views
Was this helpful?

Video transcript

Hey everyone in this problem. We are told that real gasses can approach the behavior of ideal gasses under certain conditions and that helium is thought to be the closest to an ideal gas with the ratio of heat capacities 1.66. Okay. And we're asked to calculate the molar heat capacities CV and cp of helium. All right. So, let's think about the heat capacity CV and C P K. So CV is the heat capacity with constant volume and cP is with constant pressure. We're told gamma. Okay, well, we know that gamma is going to equal to the ratio of the two. So cP over CV know that gamma is 1.661 point 66 is equal to CP over CV. Which means that we can write C P as 1. C B. Okay, so we know the relationship between the two. If we can find one of them, then we'll be able to find the other. All right, let's also recall that we can write C P is equal to C V plus R K. The gas constant. Alright, so doing this will allow us to find one of these values. Okay, Why? Well, because we can substitute what we know About the relationship here in Blue CP is equal to 1.66 CV into this other equation. Okay, so let's substitute CP is equal to 1.66 CV Into this equation with the gas constant. What we're going to find is that 1.66 CBB is equal to cv plus our I'm just gonna give ourselves some more room to work. So moving the CV to the left hand side we get 0. CV Is equal to R. Which is 8.314 eight jewels per mall Calvin dividing is gonna give us C. V. is equal to .597 jules Permal Calvin. Okay. Or if we round we get 12.6 joules per mole. Well then okay so this is gonna be our value of C. V. Now it's fine. The value of C. P. A. And I mentioned before the ratio between the two gave us the ability to calculate one as long as we found the other. Okay so now that we have Cv we can substitute that in and find cp and we're going to get Cp is equal to 1.66 times C. V. Which is 12.597. Okay, jules per mole Calvin. This is gonna give us c. p. is equal to 20.91 jules per mole. Okay and so that is our heat capacity under constant pressure. 20.91 jewels for both Calvin. And so the answer is going to be answer. C Okay. The heat capacity under constant volume is 12.6 joules per mole kelvin and under constant pressure is 20.91 joules per mole kelvin. That's it for this one. Thanks everyone for watching. See you in the next video
Related Practice
Textbook Question
A gas in a cylinder expands from a volume of 0.110 m^3 to 0.320 m^3 . Heat flows into the gas just rapidly enough to keep the pressure constant at 1.65 * 10^5 Pa during the expansion. The total heat added is 1.15 * 10^5 J. (a) Find the work done by the gas.
484
views
Textbook Question
Boiling Water at High Pressure. When water is boiled at a pressure of 2.00 atm, the heat of vaporization is 2.20 * 10^6 J/kg and the boiling point is 120°C. At this pressure, 1.00 kg of water has a volume of 1.00 * 10^-3 m^3 , and 1.00 kg of steam has a volume of 0.824 m^3. (a) Compute the work done when 1.00 kg of steam is formed at this temperature.
2198
views
Textbook Question
Heat Q flows into a monatomic ideal gas, and the volume increases while the pressure is kept constant. What fraction of the heat energy is used to do the expansion work of the gas?
616
views
Textbook Question
During an isothermal compression of an ideal gas, 410 J of heat must be removed from the gas to maintain constant temperature. How much work is done by the gas during the process?
1078
views
Textbook Question
Figure E19.8 shows a pV-diagram for an ideal gas in which its absolute temperature at b is one-fourth of its absolute temperature at a.

(d) Did heat enter or leave the gas from a to b? How do you know?

788
views
Textbook Question
The process abc shown in the pV-diagram in Fig. E19.11 involves 0.0175 mol of an ideal gas.

(a) What was the lowest temperature the gas reached in this process? Where did it occur?

1214
views
1
rank