At what temperature 28 gm of N2 will occupy a volume of 20 litres at 2 atm?
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Hey, here is ur answer ...
Given, w = 28gm, M = 28 gm (since, mass of N2 is 28gm) , V = 20lit. , P = 2atm..
We know that,, PV = nRT..,(ideal gas equation)
PV = w/M × RT...(since, n = w/m)
substitute the given values in the equation...
then, (2)(20) = (28/28)(0.0821)T
= 40 = 1(0.0821)T
T = 40/0.0821
T = 487.21 Kelvin...
Therefore, temperature is 487.2K or 214.2°C...
hope this helps u....
plzz mark it as brainliest.....
Given, w = 28gm, M = 28 gm (since, mass of N2 is 28gm) , V = 20lit. , P = 2atm..
We know that,, PV = nRT..,(ideal gas equation)
PV = w/M × RT...(since, n = w/m)
substitute the given values in the equation...
then, (2)(20) = (28/28)(0.0821)T
= 40 = 1(0.0821)T
T = 40/0.0821
T = 487.21 Kelvin...
Therefore, temperature is 487.2K or 214.2°C...
hope this helps u....
plzz mark it as brainliest.....
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