Question

A litre of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure,if the pressure is made 75 atm then at which temperature will 1 litre of the same gas weigh 1 gram? Solution=?

Answer 1

45000 K .

Step-by-step 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 :

`PV=nRT`

Here , n is no of moles ,
`n=(Given \ Weight )/(Molecular\ Mass)=(w)/(M)`

Putting initial and final values and dividing them :

`(P_1V_1)/(P_2V_2)=((w_1)/(M)T_1)/((w_2)/(M)T_2)`

`(1* 1)/(75* 1)=((2)/(M)* 300)/((1)/(M)* T_2)\\ \\T_2=45000\ K.`

Hence , this is the required solution.