Answer:
45000 K .
Explanation:
Given :
A liter of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure
We need to find the temperature in which 1 litre of the same gas weigh 1 gram
in pressure 75 atm.
We know, by ideal gas equation :
[tex]PV=nRT[/tex]
Here , n is no of moles , [tex]n=\dfrac{Given \ Weight }{Molecular\ Mass}=\dfrac{w}{M}[/tex]
Putting initial and final values and dividing them :
[tex]\dfrac{P_1V_1}{P_2V_2}=\dfrac{\dfrac{w_1}{M}T_1}{\dfrac{w_2}{M}T_2}[/tex]
[tex]\dfrac{1\times 1}{75\times 1}=\dfrac{\dfrac{2}{M}\times 300}{\dfrac{1}{M}\times T_2}\\ \\T_2=45000\ K.[/tex]
Hence , this is the required solution.
