Chemistry --- Gas Law Problem?

Question

A certain volume of oxygen at an initial pressure of 101.3 kPa is compressed to four-fifths of its original volume. What is the new pressure?

Sorry i have no idea how to figure this out because I dont know what equation to use. Thanks.

Answer 1

Boyle's law: when temp. remains constant, P1*V1=P2*V2

where P1 and V1 are the initial pressure and volume
And P2 and V2 are the final pressure and volume

Therefore,

101.3*V1=P2*4/5*V1.......(V2=4/5 of V1)

101.3*V1=P2*4V1/5

101.3*V1*5=P2*4V1

(506.5*V1)/4V1=P2

(506.5)/4=P2

.`.P2=126.625kPa

Answer 2

The question seems to make the assumption that the tempurature remains constant.

Pressure X Volume should = Pressure X Volume if the other side of the PV=nRT equation remains constant.

101.3 x V = newP x .8V

newP = 101.3/.8 = 126.625 kPa.

Answer 3

Treating it as an ideal gas, you can use PV/T = constant.

Assuming temperature is also constant leaves us with PV = constant.

Thus if V -> 4/5 V

Then P -> 5/4 P

The new pressure is (5/4)*101.3 kPa = 126.6 kPa

Answer 4

artwork out the molecular weights for CH4 and 2 O2 be conscious that they occupy an area (quantity) at 2 atms whilst the reaction happens there is largely one gas final and you will possibly be able to desire to nicely known the molecular weight of this gas examine Avagadros quantity (to get an expertise) and use P1V1/T1 = P2V2/T2 or PV=RT (T is in Kelvin)

Answer 5

PiVi=PfVf
Pi=initial pressure
Vi=initial volume
Pf=final pressure
Vf=final volume
101,3kPa.V =Pf.4/5V
Pf=126,625kPa

Answer 6

(P1V1)/T1 = (P2V2)/T2
Since T is constant,
P1V1 = P2V2

(P1V1)/V2 = P2

P1 = 101.3
V1 = 1
V2 = 4/5 or 0.8

I'll let you do the math.