Chemistry help involving temp, pressure and liters?

All right this is simple.

You can just use the formule P1V1/T1 = P2V2/T2

In the problem you already have P1, V1, T1, V2, and T2. So you are finding P2.

So you manipulate the formula so you are solving for P2.

P2 = P1V1T2 / T1V2

Then you have to change 762 torr to atm and 26.9 degrees celsius to Kelvin.

There are 760 torr in 1 atm

So you divide 762 torr by 760

762 torr * 1 atm / 760 torr = 1.00 atm

Now in order to change Celsius to Kelvin you just add 273

26.9 + 273 = 299.9K

Now you just plug in the numbers.

P2 = (1 atm)(1.40L)(299.9K)/(299.9K)(0.150L) = 9.33 atm

Hope that helped!!

Variables:

V1 = 1.40 L

P1 = 762 torr

V2 = 0.15 L

P2 = ? atm

Temperature can be ignored because it is constant.

Since your problem deals with volume and pressure only, use Boyle's Law:

P1*V1 = P2*V2

Solve for the missing variable:

P2 = P1*V1 / V2

P2 = (762 torr * 1.40 L) / (0.150 L)

P2 = 7112 torr

Then convert this to the proper unit and round to three significant figures.

1 atm = 760 torr

7112 torr * (1 atm / 760 torr) = 9.36 atm

Make sure your answer makes sense. Your initial pressure was basically 1 atm. Then you reduced the volume by nearly a factor of ten. The pressure then must increase by nearly a factor of ten, and it does.

Since temperature remains constant, we will apply Boyle's law

P1 V1 = P2 V2 P1= 762 torr V1= 1.40L V2= 0.150L P2=?

P2 = P1V1/ V2 = (762 X 1.40) / 0.150 = 7112 torr =9.36 atm

Looks like its a lot of pressure hmm......

Use the combined gas law to find the ratio of the final V to the initial V. P1V1 / T1 = P2V2 / T2 720. torr x V1 / 298 K = 605 torr x V2 / 288 K V2 / V1 = (720. torr x 288 K) / (605 torr x 298 K) = 1.15 The final volume will be 1.15 times greater than the original volume. The change will be 0.15 of the original (or 15 % greater).

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I just taught this in chemistry today. I have a simple method to performing all this.