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# Need help on Gas equation question

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1. Calculate the volume in Dm^3 occupied by 0.10mol of carbon dioxide at 1.01x10^5Nm^-2 and 110 degrees C.(gas constant R = 8.31JK^-1mol^-1) please could someone help me with this , I just can't seem to get the right answer ? The answer is 3.15dm^3
2. As pV=nRT
p = 1.01x10^5
so V = (nRT)/p
so temperature is 383.15K
putting this into the equation will return 0.00315m^3
as the relationship is cubic, m^3 --> dm^3 is m^3 * 10^3
3. .
just multiply this answer by 1000 to convert the volume from m3 to dm3 .
4. (Original post by rebeccaslugocki)
Calculate the volume in Dm^3 occupied by 0.10mol of carbon dioxide at 1.01x10^5Nm^-2 and 110 degrees C.(gas constant R = 8.31JK^-1mol^-1) please could someone help me with this , I just can't seem to get the right answer ? The answer is 3.15dm^3
pV = nRT

First, convert all the given values into the correct units required for the formula:
110 degrees = 383K (110+273)
We only need to convert the temperature, all the other units are fine.

Now we rearrange the formula for what we are trying to find, which in this case is V (volume):

V = nRT / p

Substitute in the values:

0.10 x 8.31 x 383 / 1.01x10^5 = 0.003151 (this is in the default unit which is in metres cubed)

Convert from metres cubed to cm cubed:

0.003151 x 1,000,000 = 3151 cm^3
Now, convert from cm cubed to dm cubed:

3151 / 1000 = 3.15 dm^3
5. (Original post by derpz)
pV = nRT

First, convert all the given values into the correct units required for the formula:
110 degrees = 383K (110+273)
We only need to convert the temperature, all the other units are fine.

Now we rearrange the formula for what we are trying to find, which in this case is V (volume):

V = nRT / p

Substitute in the values:

0.10 x 8.31 x 383 / 1.01x10^5 = 0.003151 (this is in the default unit which is in metres cubed)

Convert from metres cubed to cm cubed:

0.003151 x 1,000,000 = 3151 cm^3
Now, convert from cm cubed to dm cubed:

3151 / 1000 = 3.15 dm^3
Thank you ! Turns out I was multiplying by 273 instead of adding and I didn't know what the default was either. Thanks

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