Question

Assume that the pressure in a room remains constant at 1.01

x10^5 Pa and the aire is composed only of nitrogen. The volumeof
the room is 60.0 m^3. When the temperature increases from 289 to302
K, what mass of air (in kg) escapes from the room?

Answer 1

Step-by-step explanation:

Given

Pressure
`P_1=1.01* 10^5 Pa`

Volume of air
`V=60 m^3`

Initial Temperature
`T_1=289 K`

`T_2=302 K`

Initial moles is given by

`PV=nRT`

`n_1=(P_1V_1)/(RT_1)`

`n_1=(1.01* 10^5* 60)/(R\cdot 289)`

when some gas escape out

no of moles is equal to

`n_2=(P_2* V_2)/(R\cdot T_2)`

`n_2=(1.01* 10^5* 60)/(R\cdot 302)`

remaining no of moles
`=n_1-n_2`

`=(1.01* 10^5* 60)/(R\cdot 289)-(1.01* 10^5* 60)/(R\cdot 302)`

`=(1.01* 10^5* 60)/(R)((1)/(289)-(1)/(302))`

Mass of air escape out

`=(n_1-n_2)* M`

`=((1.01* 10^5* 60)/(R)((1)/(289)-(1)/(302)))* 28`

`=25.272* 10^3\ g`

`m=25.272\ kg`