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Chemistry 14 Online
OpenStudy (anonymous):

How would I do this? Use Boyle's law to calculate the volume the gas will occupy at 23C and pressure is increased to 1.55 atm.

OpenStudy (anonymous):

I did this: \(\ \Huge V=k \times \frac{1}{p} ,\) \[\ \Huge = 23 \times \frac{1}{1.55}, \] \[\ \Huge So, V\approx14.8L. \] \[\ \Huge However, \]\ \[\ \Huge The correct answer is 6.54 L... \]

OpenStudy (anonymous):

23 C isnt that the temperature? also boyle's law k is a constant the equation should be pV=pV

OpenStudy (anonymous):

what was the initial volume and initial pressure?

OpenStudy (anonymous):

So temperature doesn't have anything to do with the equation?

OpenStudy (anonymous):

The inital temperature was 23C, and the inital pressure was 748 torr.

OpenStudy (anonymous):

gosh i have lousy memory i believe 1atm was 780 mm Hg or 780 torrs, you might wanna double check that so initial pressure would be 748/780 the initial volume should be given to you and temperature wont affect it unless the temperature is changing ideal gas law PV=nRT

OpenStudy (anonymous):

How would I set up this eqn?

OpenStudy (anonymous):

assuming the amount of gas you're using is constant and temperature is constant PV=PV

OpenStudy (anonymous):

I still can't get the correct answer...

OpenStudy (anonymous):

can you type the full question?

OpenStudy (anonymous):

A fixed quanitity of gas at 23C exhibits a pressure of 748 torr and occupies a volume of 10.3L . a) Use Boyle's law to calculate the volume the gas will occupy at 23C if the pressure is increased to 1.55 atm, and b) Use Charles's law to calculate the volume the gas will occupy if the temperature is increased to 145C while the pressure is held constant

OpenStudy (anonymous):

1 atm= 760 torr given the answer the initial volume of would be 10.3 L look at your question to see if this number appears anywhere (748/760)V=1.55(6.54) V=10.3

OpenStudy (anonymous):

initial volume is 10.3L as problem state initial pressure is 748 torr= 748/760 atm using boyle law PV=PV (10.3L)(748/760 atm)=V(1.55 atm) V=[10.3(.9842 atm/1.55 atm)]L

OpenStudy (anonymous):

Where does the 760 come from?

OpenStudy (anonymous):

760 torr = 1atm unit conversion

OpenStudy (anonymous):

look it up

OpenStudy (anonymous):

Oh okay, so in the formula, the left side is inital, and the right side is according to (a)

OpenStudy (anonymous):

Boyle's Law is basically PV=PV?

OpenStudy (anonymous):

problem 2, temperature is increasing and pressure remains constant PV=nRT V/T= nR/P because nR/P is just a constant value, V/T=V/T

OpenStudy (anonymous):

yes boyles law is PV=PV charles law i assume is V/T=V/T

OpenStudy (anonymous):

This makes sense now, thanks for all your help!

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