The given condition is STP, under this condition, gas has a rule of 22.4 L per mole. And the given equation is already balanced. The ratio of mole number is the same as the ratio of the volume and is also the same as the ratio of coefficients. So the answer is 4.0 liters.
Answer:
Humans have impacted the hydrosphere drastically and will only continue to due so based on population needs. Global climate change, water pollution, damming of rivers, wetland drainage, reduction in stream flow, and irrigation have all exerted pressure on the hydrosphere's existing freshwater systems.
Hope it helps :)
Answer:
![PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]](https://tex.z-dn.net/?f=PV_%7Bm%7D%20%3D%20RT%5B1%20%2B%20%28b-%5Cfrac%7Ba%7D%7BRT%7D%29%5Cfrac%7B1%7D%7BV_%7Bm%7D%20%7D%20%2B%20%5Cfrac%7Bb%5E%7B2%7D%20%7D%7BV%5E%7B2%7D%20_%7Bm%7D%20%7D%20%2B%20...%5D)
B = b -a/RT
C = b^2
a = 1.263 atm*L^2/mol^2
b = 0.03464 L/mol
Explanation:
In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:
Using the van deer Waals equation of state:

With further simplification, we have:
![P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]](https://tex.z-dn.net/?f=P%20%3D%20RT%5B%5Cfrac%7B1%7D%7BV_%7Bm%7D-b%20%7D%20-%20%5Cfrac%7Ba%7D%7BRTV_%7Bm%7D%20%5E%7B2%7D%20%7D%5D)
Then, we have:
![P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BRT%7D%7BV_%7Bm%7D%20%7D%20%5B%5Cfrac%7B1%7D%7B1-%5Cfrac%7Bb%7D%7BV_%7Bm%7D%20%7D%20%7D%20-%20%5Cfrac%7Ba%7D%7BRTV_%7Bm%7D%20%7D%5D)
Therefore,
![PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]](https://tex.z-dn.net/?f=PV_%7Bm%7D%20%3D%20RT%5B%281-%5Cfrac%7Bb%7D%7BV_%7Bm%7D%20%7D%29%20%5E%7B-1%7D%20-%20%5Cfrac%7Ba%7D%7BRTV_%7Bm%7D%20%7D%5D)
Using the expansion:

Therefore,
![PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]](https://tex.z-dn.net/?f=PV_%7Bm%7D%20%3D%20RT%5B1%2B%5Cfrac%7Bb%7D%7BV_%7Bm%7D%20%7D%2B%5Cfrac%7Bb%5E%7B2%7D%20%7D%7BV_%7Bm%7D%20%5E%7B2%7D%20%7D%20%2B%20...%20-%5Cfrac%7Ba%7D%7BRTV_%7Bm%7D%20%7D%5D)
Thus:
equation (1)
Using the virial equation of state:
![P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]](https://tex.z-dn.net/?f=P%20%3D%20RT%5B%5Cfrac%7B1%7D%7BV_%7Bm%7D%20%7D%2B%20%5Cfrac%7BB%7D%7BV_%7Bm%7D%20%5E%7B2%7D%7D%2B%5Cfrac%7BC%7D%7BV_%7Bm%7D%20%5E%7B3%7D%20%7D%2B%20...%5D)
Thus:
equation (2)
Comparing equations (1) and (2), we have:
B = b -a/RT
C = b^2
Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.
[/tex] = 0.03464 L/mol
a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2
Answer:
773.51495 grams
Explanation:
1 moles KBr to grams = 119.0023 grams
6.5*119.0023 = 773.51495 grams