Question

At what temperature (in K) does NH₃ have a density of 0.812 g/L at 1.17 atm?

Answer 1

299

Step-by-step explanation:

Or 298.51 to be exact!

Answer 2

The temperature is "298.51 K".

Step-by-step explanation:

The given values are:

Density of ammonia,

d = 0.812 g/L

Pressure,

P = 1.17 atm

Mass:

M =
`NH_3`

=
`14+ 1* 3`

=
`17`

As we know,

⇒
`PV=nRT`

∴
`T=(PV)/(nR)`

`=(PV)/((W)/(M)* R )`

`=(PVM)/(W* R)`

By putting the value of "W", get

`=(PVM)/(V* d* R)`

`=(P* M)/(d* R)`

On substituting the values in the above equation, we get

`=(1.17* 17)/(0.812* 0.08205)`

`=(19.89)/(0.06699)`

`=298.51 \ K`