Assume·ll reactantsand products are gases @ STT.C, 1 atms.) 22.41/mol of any g h

Question

Assume·ll reactantsand products are gases @ STT.C, 1 atms.) 22.41/mol of any g hemistry: Gas La Pressure / Boyle's Law/Charles Law Valevie kun ALL WORK or no credit will be assigned h unless otherwise indicated. SHOW A units of pressure correponding to 1 atmosphere of pressure. O pts) F Fitt in che table with the three atm PS I e correct answen. a PtJ Going down a mountain air flows (inout) the eustachian tube. (Circle th (1 pes.) write the formula for the relationship between pressure and volume a ptJ What is the name of the law for the relationship between 45 pts.) The pressure on a 2.00 liter . V1 VJ 4 S. S pts) The press are on a 2,00 titer bailoon went from 1.00 atm of pressure to 500. torr of pressure. auw 5, a pt.) Convert the final pressure to atm units. (1 pt) The volume will: increase/decrease/stay the same. (Circle the correct answer) (3 pts) Calculate the final volume. 6. (4 pts) A balloon at 0.800 L of volume and 3.00 atm expanded to 5.00 L of volume. (1 pt) The pressure: increased/decreased/stayed the same? (Circle the correct answer) 3 pts) Calculate the final pressure. 7. (1 pt) Los Alamos and South Lake Tahoe are both high-altitude citie weighs slightly more than a liter of Los Alamos air. Which city has the highest altitude? LOS Alam OS 8. (1 pt.) For the Charles law to work, what temperature scale must the temperature always be in? Kelvins 9. (1 pt.) What is zero degrees in the temperature scale in number 8 called. Abso lute two

Explanation / Answer

3.

P x V = constant

Or, P1 V1 = P2 V2

4.

Boyle's law: Boyle’s states that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.

5.

(i) 500 torr

1 torr = 0.00131579 atm
or, 500 torr = 500 x 0.00131579 atm
                     = 0.658 atm

(ii) pressure decreased from 1 atm to 0.658 atm
Since, pressure is decreased, volume will increase.

(iii)
P1 = 1 atm, V1 = 2 L, P2 = 0.658 atm, V2 = ?

P2V2 = P1V1
or, V2 = P1V1 / P2
or, V2 = [1 atm) (2 L) ] / ((0.658 atm) = 3.04 L

6.

P1 = 0.800 L, P1 = 3.00 atm, V2 = 5.00 L

Since volume is increased from 0.8 L to 5 L, so pressure will decrease.

Now, we know that

P2V2 = P1V1
or, P2 = P1V1 / V2
            = (3 atm) (0.8 L) / ( 5 L)
            = 0.48 atm

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