Difference between revisions of "Ch5 Lec 3"
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:[[File:Screen Shot 2020-03-29 at 11.21.43 AM.png|300px|thumb|center|from Crash Course]] | :[[File:Screen Shot 2020-03-29 at 11.21.43 AM.png|300px|thumb|center|from Crash Course]] | ||
| − | When combining all of the gas laws, Boyles, Charles', and Avogadro's we get the ''Ideal Gas Law Equation:'' | + | When combining all of the gas laws, Boyles, Charles', and Avogadro's we get the ''Ideal Gas Law Equation:'' As with any mathematical relationship/equation, if there are 5 variables (P, V, n, R, T) all you need to know is 4 of them and the 5th can be calculated. |
| + | ''R'', the gas constant is equal to (depending on the units): | ||
| + | :[[File:Screen Shot 2020-03-30 at 2.58.32 PM.png|300px]] | ||
| + | ====Question 1==== | ||
| + | If given the following data, which gas constant would you use? | ||
| + | :P = 1.12 atm | ||
| + | :V= 2.45 L | ||
| + | :n = 1.00 moles | ||
| + | :T = 298 K | ||
| + | <div align="right">Answer: 0.082057 L*atm/(mol*K)</div> | ||
| + | If the volume was not given in the question above, show how it can be calculated: | ||
| + | :P = 1.12 atm | ||
| + | :n = 1.00 moles | ||
| + | :T = 298 K | ||
| + | :R = 0.082057 l*atm/(mol*K) | ||
| + | '''''Calculate V and make sure your work shows how the units cancel:''''' | ||
| + | ====Question 2==== | ||
| + | If given the V, n, and T, which gas constant would you use if you want to calculate the pressure in "bar"? | ||
| + | <div align="right">Answer: 0.08314 L*bar/(mol*K)</div> | ||
| + | |||
| + | Given: | ||
| + | :V= 22.414 L | ||
| + | :n = 1.0000 moles | ||
| + | :T = 273.15 K | ||
| + | :R = from above... | ||
| + | '''''Calculate the pressure in bar and make sure your work shows how the units cancel:''''' | ||
| + | |||
| + | |||
| + | Comments on Temperature: Always use Kelvin. <sup>o</sup>C + 273.15 = K | ||
| + | |||
| + | |||
| + | Nearly all gases behave similarly when under a set of conditions (P, V, and T). So much so that we define a set of conditions, "standard temperature and pressure" (STP) as T = 273.15 K and P = 1 atm. Under STP, the volume of 1.00 moles of gas is 22.4 L. This concept was discussed in the crash course video. | ||
:[[File:Screen Shot 2020-03-29 at 11.20.53 AM.png|300px|thumb|center|from Crash Course]] | :[[File:Screen Shot 2020-03-29 at 11.20.53 AM.png|300px|thumb|center|from Crash Course]] | ||
| + | |||
| + | |||
| + | ''As you might expect...there is a short WA waiting for you...(8 am, Weds)...'' | ||
| + | |||
| + | '''End of Chapter 5''' | ||
Latest revision as of 21:16, 30 March 2020
in progress...
Ideal Gas Law Equation (Sec 5.4)
When combining all of the gas laws, Boyles, Charles', and Avogadro's we get the Ideal Gas Law Equation: As with any mathematical relationship/equation, if there are 5 variables (P, V, n, R, T) all you need to know is 4 of them and the 5th can be calculated.
R, the gas constant is equal to (depending on the units):
Question 1
If given the following data, which gas constant would you use?
- P = 1.12 atm
- V= 2.45 L
- n = 1.00 moles
- T = 298 K
If the volume was not given in the question above, show how it can be calculated:
- P = 1.12 atm
- n = 1.00 moles
- T = 298 K
- R = 0.082057 l*atm/(mol*K)
Calculate V and make sure your work shows how the units cancel:
Question 2
If given the V, n, and T, which gas constant would you use if you want to calculate the pressure in "bar"?
Given:
- V= 22.414 L
- n = 1.0000 moles
- T = 273.15 K
- R = from above...
Calculate the pressure in bar and make sure your work shows how the units cancel:
Comments on Temperature: Always use Kelvin. oC + 273.15 = K
Nearly all gases behave similarly when under a set of conditions (P, V, and T). So much so that we define a set of conditions, "standard temperature and pressure" (STP) as T = 273.15 K and P = 1 atm. Under STP, the volume of 1.00 moles of gas is 22.4 L. This concept was discussed in the crash course video.
As you might expect...there is a short WA waiting for you...(8 am, Weds)...
End of Chapter 5