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Finding the temprature watch

1. If we have the volume of sample of nitrogen gas that equal to 0.080 m^3 in Standard atmosphere pressure 101 pa and there is 3.6 mol from the gas then what is the temperature of the gas

I think the law to use is :Pv=nRT .But I am not sure
2. (Original post by MAA_96)
If we have the volume of sample of nitrogen gas that equal to 0.080 m^3 in Standard atmosphere pressure 101 pa and there is 3.6 mol from the gas then what is the temperature of the gas

I think the law to use is :Pv=nRT .But I am not sure
Yep, that's the correct law (generally volume is a capital V).
3. (Original post by PythianLegume)
Yep, that's the correct law (generally volume is a capital V).
So I have p =101,V=0.80,n=3.6

And?
4. (Original post by MAA_96)
So I have p =101,V=0.80,n=3.6

And?
Isn't the standard atmospheric pressure (approx.) 1.0 x 10^5 Pa?

In which case you would do:

pV = nRT

therefore:

T = pV/nR = (1.0 x 10^5 x 0.080)/(3.6 x 8.31)

Plug that into your calculator, and you should get a value for the temperature in kelvin.

(Btw you have given two different volumes, not sure which you meant but I used the first)

Posted from TSR Mobile
5. (Original post by Rhodopsin94)
Isn't the standard atmospheric pressure (approx.) 1.0 x 10^5 Pa?

In which case you would do:

pV = nRT

therefore:

T = pV/nR = (1.0 x 10^5 x 0.080)/(3.6 x 8.31)

Plug that into your calculator, and you should get a value for the temperature in kelvin.

(Btw you have given two different volumes, not sure which you meant but I used the first)

Posted from TSR Mobile
Thank you and yes I meant the first volume

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