Gas problem

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Dencology

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if a balloon initially has a P=27 atm at 250K and the temperature is raised to 300K, what chage in quantity of gas present must be made to maintain the pressure of the system assuming volume remains fixed?

a. 20% decrease in mole
b. 20% increase in mole
c. 25% increase ""
d. 33% """"
e. none of the above.

can anyone explain?
ans. E.

2. a solution of KNO3 is known to have a 0.564 molal conc. to calculate the concentration of this solution in terms of molarity, which of the following must be specified?

a. density of the solvent
b. volume
c. temp
d. solubility

i was not sure between a & b. the ans. is a. any help?

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lol ok enough jokes. here's how i think you solve the first question. Since volume is staying the same in both problems, create the two equations equal to volume and then combine them

PV=n1RT -> 27v = .08(250)n -> v = (.08(250)n1)/27

PV = n2RT -> 27v = .08(300)n -> v = (.08(300)n2)/27

equal them to each other and you get

(.08(250)n1)/27 = (.08(300)n2)/27

you can really cancel all the common factors in both problems thereby just leaving you with 250n1 = 300n2

divide n2 by 300 and simplify and you get (5/6)n1 = n2

meaning n2 is about 70% of n1 about. (decrease of 30%)

I guess if you think about it logically, since temp is increasing in the second one, and everything's remaining constant, the moles has to compensate for that by being lower. So you can cancel out D and E. And then the equation gives you how much of a decrease...
 
for the other question:

If you have .564 molality is basically .564 moles/Kg

So you would have to know density which is g/L to figure out molarity because if you multiple .564moles/kg by kg/L you will get moles/L which is molarity.
 
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if a balloon initially has a P=27 atm at 250K and the temperature is raised to 300K, what chage in quantity of gas present must be made to maintain the pressure of the system assuming volume remains fixed?

a. 20% decrease in mole
b. 20% increase in mole
c. 25% increase ""
d. 33% """"
e. none of the above.

can anyone explain?
ans. E.

2. a solution of KNO3 is known to have a 0.564 molal conc. to calculate the concentration of this solution in terms of molarity, which of the following must be specified?

a. density of the solvent
b. volume
c. temp
d. solubility

i was not sure between a & b. the ans. is a. any help?

Question1/
I am not sure if i am right. Please correct me.
The variables is only n, T since V,P,R are constant-->
n1T1=n2T2=k
--> n2=n1*(T1/T2)=(250/300)n2=0.83(n1).--> Thus, n2 is 83% n1
OR if you think about Le Chateller principle:
PV=nRT
Temperature increase will lead to Pressure increase. To maintain the same pressure, V must increase. If V must increase, the mole products n2 must increase.
 
If Temp. increases so will Volume at constant P (charles's Law).
Then if at constant Pressure moles will also increase if Volume increases. (avoCado's Law)
 
please tell me if this is right...if the volume remains constant as it says so in the problem, then the number of moles must remain constant as well, thus explaining why the answer was e. no calculations are needed right????????????????????????
 
uhh i don't see how that makes sense. Why would volume remaining constant make the number of moles constant as well...read the above posts. they might clarify for you...
 
so you are saying that the n must be increased by 83% in order to satisfy the question?


Question1/
I am not sure if i am right. Please correct me.
The variables is only n, T since V,P,R are constant-->
n1T1=n2T2=k
--> n2=n1*(T1/T2)=(250/300)n2=0.83(n1).--> Thus, n2 is 83% n1
OR if you think about Le Chateller principle:
PV=nRT
Temperature increase will lead to Pressure increase. To maintain the same pressure, V must increase. If V must increase, the mole products n2 must increase.
 
even though volume remains the same, you're still altering the temperature. The only way to keep pressure the same while altering the temperature is to change the number of moles

just look at the PV = nRT formula...
you want to keep P the same and V is constant. value of R doesn't change...Thereby if you're increasing T, you have to decrease n by a specific amount. The only way to do that is to set the volume of 1 equal to volume of another
 
i guess we are not still sure of the answer. this is a concept question which could very well be like one on real DAT. anyone out there that can explain?
 
Use PV=nRT, everything is the same except for temperature and the number of moles, which is the point of the question.

n1: 27V/R(250)
n2: 27V/R(300)

Find the ratio of n1 and n2. You basically get 250/300 = n1 / n2. So that means 250 n2 equals 300 n1. This is equivalent to 5 n2 = 6 n1 or n1=5/6 n2. If you had 100 moles of n1 (the original number of moles) and the temp was increased, then you will have 100*(5/6) moles of n2, which is 83.33. This is a 100-83.33 = 17, a 17% decrease.
 
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