How to determine the pressure when you dont have the volume

[arhzz]

Homework Statement

150ml of air under a pressure of 1013hPa are sucked into a bicycle pump by pulling out the piston. The opening of the pump is then sealed airtight with a stopper. Now push the plunger in halfway, then further in up to 3/4.
a) How great is the pressure if the temperature remains the same?
b) In the case of an isobaric implementation - how far would the temperature change compared to the initial temperature?

Relevant Equations

p1V1 = p2V2

Hello! I am having trouble with this problem I found online,it was listed under the "easy" category yet I am somehow not seeing the trick.

First I converted the V into m^3 so that V = 0,015 m^3,than I converted the hPa into Pa so p = 101300 Pa and now I used the fact that ##p1V1 = p2V2 ## We can get p2 out of this equation. $$ p2 = \frac{p1V1}{V2} $$ Now the problem is I don't have V2; I am pretty much certain the problem has/can be solved this way but I'm kind of not getting it.

Thank you!

 

[Master1022]

Hi there.

Now push the plunger in halfway, then further in up to 3/4.

Can we not just use this ratio of the volumes without knowing the exact numbers? That is, we are told that we first push the plunger in half the way, which suggests that we now have a volume ## \frac{1}{2} V_1 ##. Then we push it into 3/4, so we have ## \frac{1}{4} V ## left?

I hope that is of some help.

 

[arhzz]

Yea I've actually figured it out just this morning.The question was really simple but I was too tired to think straight and after a good night sleep I solved it without issue. Here is the solution;

a) $$ p2 = \frac{p1V1}{1/2V1} $$ p2 should come out to ## p2 = 2p1 ## and for 3/4 ## p2 = 4pi ##

b) The same trick just with the temperature just that the relationship is diffrent $$ \frac{V1}{T1} = \frac{V2}{T2}$$ for 1/2 T2 should be ## T2 = 1/2 T1 ## and for 3/4 ## T2 = 1/4 T1 ##

Still thank you for your help!