Calculation of a volume given its gass mass
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Calculation of a volume given its gass massCalculate the volume of 2g of C02 (how do i work it out step by step?)
Heres my attempt of it: n=m/M so 2/44 =0.045mol So 0.045 x 24000 = 1080cm3?
Help, I got a test on friday and if i fail it am moved to btec ='[ -
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Re: Calculation of a volume given its gass mass(Original post by lilyobz)
Calculate the volume of 2g of C02 (how do i work it out step by step?)
Heres my attempt of it: n=m/M so 2/44 =0.045mol So 0.045 x 24000 = 1080cm3?
Help, I got a test on friday and if i fail it am moved to btec ='[
PV = nRT
You want to find V therefore you're looking at:
V = nRT/P
You therefore need to know:
- R (gas constant) = 8.314
- T (temperature?) = ???
- P (pressure?) = ???
- n (moles of gas), which as you've correctly worked out as 0.045mol.
Therefore:
- T = 25°C, id est 298.15K
- P = 100kPa
V = nRT/P
V = (0.045*8.314*298.15)/100
V = 111.5468/100
V = 1.116L
(which if you wanted it in cubic metres, is 1.116dm3 ≡ 1116cm3)
Though it has been at least 2 years since I last did this, so don't take my word for it...!Last edited by Friar Chris; 12-09-2012 at 20:39. -
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Re: Calculation of a volume given its gass massYour method is correct but you rounded prematurely - keep m/M in your calculator.(Original post by lilyobz)
Calculate the volume of 2g of C02 (how do i work it out step by step?)
Heres my attempt of it: n=m/M so 2/44 =0.045mol So 0.045 x 24000 = 1080cm3?
Help, I got a test on friday and if i fail it am moved to btec ='[ -
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Re: Calculation of a volume given its gass mass1 mol is 24000cm^3
you have 2g. 1 mol of CO2 is . . .
. . . 44g
So you have 2/44 = x of a mol
so x *24000 = your answer . . . . (I'm not doing everything for you.)Last edited by Clare~Bear; 13-09-2012 at 08:34. -
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Re: Calculation of a volume given its gass massSorry that doesn't make much sense; are T and P therefore to be worked out from this?(Original post by lilyobz)
Assume the room tempresure and pressure of the 1mol of gas has a volume of 24000com3
i.e.
V = nRT/P
∴ 24 = 1*8.314T/P
Thus 1mol of gas is 24dm3 in volume. You're trying to find 0.0454545mol (2/44) of gas, so it's simply:
0.04545...*8.314*T/P = 0.04545*24
0.04545...*24 = 1.090909090...
0.04545...*24 = 1.091dm3 ≡ 1091 cm3
As you said.
What possessed you to write the original question out without these data?! I went through an entire PV=nRT just to find it was a simple matter of working out 4.5% of a given value!
x*24000, not divide(Original post by Clare~Bear)
1 mol is 24000cm^3
you have 2g. 1 mol of CO2 is . . . . . . 44g
So you have 2/44 = x of a mol
so x /24000 = your answer . . . . (I'm not doing everything for you.)
But why OP didn't add the 24dm3 information in his original post... who knows.Last edited by Friar Chris; 12-09-2012 at 21:02. -
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Re: Calculation of a volume given its gass masssorry(Original post by Friar Chris)
Sorry that doesn't make much sense; are T and P therefore to be worked out from this?
i.e.
V = nRT/P
∴ 24 = 1*8.314T/P
Thus 1mol of gas is 24dm3 in volume. You're trying to find 0.045mol of gas, so it's simply:
0.045*8.314*T/P = 0.045*24
0.045*24 = 1.08dm3 ≡ 1080 cm3
As you said.
What possessed you to write the original question out without these data?! I went through an entire PV=nRT just to find it was a simple matter of working out 4.5% of a given value!
x*24000, not divide
But why OP didn't add the 24dm3 information in his original post... who knows.
, the 0.45 was recurring and I rounded up prematurely there.
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Re: Calculation of a volume given its gass massI've never seen that first expression in my life. What is each constituent letter?(Original post by Friar Chris)
Sorry that doesn't make much sense; are T and P therefore to be worked out from this?
i.e.
V = nRT/P
∴ 24 = 1*8.314T/P
Thus 1mol of gas is 24dm3 in volume. You're trying to find 0.045mol of gas, so it's simply:
0.045*8.314*T/P = 0.045*24
0.045*24 = 1.08dm3 ≡ 1080 cm3
As you said.
What possessed you to write the original question out without these data?! I went through an entire PV=nRT just to find it was a simple matter of working out 4.5% of a given value!
x*24000, not divide
But why OP didn't add the 24dm3 information in his original post... who knows. -
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Re: Calculation of a volume given its gass massisnt it sopose to be x TIMES 24000 = my answer?(Original post by Clare~Bear)
1 mol is 24000cm^3
you have 2g. 1 mol of CO2 is . . . . . . 44g
So you have 2/44 = x of a mol
so x /24000 = your answer . . . . (I'm not doing everything for you.) -
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Re: Calculation of a volume given its gass massI don't think any of the gas laws or the Ideal gas equation are part of A-Level or GCSE chem courses. The constant 24dm3 for Avogadro's law at RTP is to be known, but not in name.(Original post by Friar Chris)
Sorry that doesn't make much sense; are T and P therefore to be worked out from this?
i.e.
V = nRT/P
∴ 24 = 1*8.314T/P
Thus 1mol of gas is 24dm3 in volume. You're trying to find 0.0454545mol (2/44) of gas, so it's simply:
0.04545...*8.314*T/P = 0.04545*24
0.04545...*24 = 1.090909090...
0.04545...*24 = 1.091dm3 ≡ 1091 cm3
As you said.
What possessed you to write the original question out without these data?! I went through an entire PV=nRT just to find it was a simple matter of working out 4.5% of a given value!
x*24000, not divide
But why OP didn't add the 24dm3 information in his original post... who knows. -
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Re: Calculation of a volume given its gass mass(Original post by fletchdd02)
I've never seen that first expression in my life. What is each constituent letter?
It's derived from the Ideal Gas Equation:
PV=nRT
- P is pressure (in kPa)
- V is volume (in dm3, i.e. litres)
- n is moles of substance, as always.
- R is the Gas Constant, which is 8.314
- T is temperature, in Kelvin (!! not degrees centrigrade !!)
From this therefore it can, like any equation, be re-arranged to find desired variables so long as the others are known or their values discernible from a second reading:
To find the volume, V: V = nRT/P
To find the pressure, P: P = nRT/V
To find the amount of substance: n = PV/RT
Et vice versa.
Seriously!?!?!??!(Original post by Killjoy-)
I don't think any of the gas laws or the Ideal gas equation are part of A-Level or GCSE chem courses. The constant 24dm3 for Avogadro's law at RTP is to be known, but not in name.
It's only been two years since I did my chemistry A-level, and it most definitely included this, and this was among the more basic of calculations we had to do.
In fact, I think this was in my GCSE.
It's really, really simple stuff, if they've dumbed this out of chemistry, I'd be amazed.
Yes, it is. Also I edited it to be the correct 1.091dm3, but you quoted me before I managed to submit ;D(Original post by lilyobz)
isnt it sopose to be x TIMES 24000 = my answer? -
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Re: Calculation of a volume given its gass massGas laws was removed from and replaced with environment chemistry, (Labour did this.. green hippies =/)(Original post by Friar Chris)
It's derived from the Ideal Gas Equation:
PV=nRT
- P is pressure (in kPa)
- V is volume (in dm3, i.e. litres)
- n is moles of substance, as always.
- R is the Gas Constant, which is 8.314
- T is temperature, in Kelvin (!! not degrees centrigrade !!)
From this therefore it can, like any equation, be re-arranged to find desired variables so long as the others are known or their values discernible from a second reading:
To find the volume, V: V = nRT/P
To find the pressure, P: P = nRT/V
To find the amount of substance: n = PV/RT
Et vice versa.
Seriously!?!?!??!
It's only been two years since I did my chemistry A-level, and it most definitely included this, and this was among the more basic of calculations we had to do.
In fact, I think this was in my GCSE.
It's really, really simple stuff, if they've dumbed this out of chemistry, I'd be amazed.
Yes, it is. Also I edited it to be the correct 1.091dm3, but you quoted me before I managed to submit ;D -
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Re: Calculation of a volume given its gass massWe had environment chemistry ((Original post by lilyobz)
Gas laws was removed from and replaced with environment chemistry, (Labour did this.. green hippies =/)
) back then too; it was 2010 and Labour were in power up until the May when I was in Upper Sixth, but still had gas law in there...
Besides, no reason not to teach people it on TSR. Better discussing valid chemistry here instead of try-hard indoctrination. -
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Re: Calculation of a volume given its gass massI agree with you. I encountered it in A2 Physics only.(Original post by Friar Chris)
Seriously!?!?!??!
It's only been two years since I did my chemistry A-level, and it most definitely included this, and this was among the more basic of calculations we had to do.
In fact, I think this was in my GCSE.
It's really, really simple stuff, if they've dumbed this out of chemistry, I'd be amazed.
, the 0.45 was recurring and I rounded up prematurely there.
) back then too; it was 2010 and Labour were in power up until the May when I was in Upper Sixth, but still had gas law in there...
