Question

The pressure in a container is 765.8 mmHg and the temperature is 36.4 degrees C and the volume needs a total of 25.4 cc (cm^3).The temperature rises to 63.0 deg C and the pressure changes to 9.01 atm.Convert pressure to mmHg and determine the new volume?

Answer 1

Approximately 324.1811L

Explanation:

I would use the ideal gas law `PV=nRT`, where p stands for the pressure, v stands for the volume, n stands for moles, r is the gas constant, and t stands for temperature.

We want to convert the pressure to mmHg as instructed in the problem. The conversion factor is as follows 760mmHg =1atm. So we take 9.01atm from the problem and multiply it by 760. This gives us approximately 6847.5996 mmHg.

Then convert all the units to units that will cancel with the gas constant (0.0820574`(L*atm)/(K*mol)`).

All pressure should be in atm so convert 765.8mmHg to atm using the above conversion factor of 760mmHg=1atm. We get that 765.8mmHg is equal to approximately 1.00763atm. Next, convert 25.4cc to L using the conversion factor 1cc=0.001L. We get that 25.4cc is equal to approximately 0.0254L. Finally, convert 36.4 and 63.0 degrees Celsius to Kelvin by using the conversion factor 1 Celsius = 274.15K. We get the 36.4 is equal to 309.55K and 63.0 is equal to 336.15K.

Let's plug in the first set of information into the ideal gas law.
(1.00763atm)(0.0254L)=n(0.0820574`(L*atm)/(K*mol)`)(309.55K)

Solve for n and we get approximately .0010075956 moles

Now let's plug in the changed information to solve for V
(9.01atm)V=(.0010075956 moles)(0.0820574`(L*atm)/(K*mol)`)(336.15K)

Solve for V and we get approximately 324.1811L

If you want the answer in cc you can convert it using the conversion factor 1cc=0.001L to receive 324181.1cc

Answer 2

Well, `(P_1V_1)/T_1=(P_2V_2)/T_2`

Explanation:

`V_2=(P_1V_1)/T_1xxT_2/P_2`

And the use of `mm*Hg` for a pressure OVER `1*atm` is clearly wrong-headed. And you really should not be using using `mm*Hg` as a unit of pressure UNLESS you are using a mercury manometer in your lab.

`P_1=(765.8*mm*Hg)/(760*mm*Hg*atm^-1)=1.008*atm`

And of course we report temperature in `"degrees Kelvin..."`

And so.....

`V_2=(P_1V_1)/T_1xxT_2/P_2=(1.008*atmxx25.4*cm^3xx336.2*K)/(309.6*Kxx9.01*atm)`

`=3.10*cm^3`