At room temperature, the following reaction proceeds nearly to completion:
`2NO+O_(2)to2NO_(2)toN_(2)O_(4)`
The dimer, `N_(2)O_(4)`, solidfies at `262 K`. A `250 mL` flask and a `100 mL` flask are separated by a stopcock. At `300 K`, the nitric oxide in the larger flask exerts a pressure of `1.053 atm` and the smaller one contains oxygen at `0.789 atm`. The gase are mixed by opening the stopcock and after the end of the reaction the flasks are cooled to `220 K`. Neglecting the vapour pressure of the dimer, find out the pressure and composition of the gas remaining at `220 K`. (Assume the gases to behave ideally)

Explanation

`PV=nRT`
or `n=(PV)/(RT)`
Moles of `NO`
`a=(1.053xx250)/(0.0821xx300xx1000)=0.01069`
Moles of `O_(2)`
`b=(0.789xx100)/(0.0821xx300xx1000)=0.003203`
Number of moles of oxygen is very small, so it must be the limiting reagent.
`underset((a-2b))underset(a)(2NO)+underset(0)underset(b)(O_(2))to2NO_(2)tounderset(b)underset(0)(N_(2)O_(4))`
All oxygen is consumed.
Moles of `NO` left`=(a-2b)`
`=0.01069-2(0.003203)`
`=0.004284`
At `220 K`, the dimer is solidified and it does not exert any pressure (neglecting vapour pressure).
All the pressure is due to the left `NO` only.
`PV=nRT`
or `P=(n)/(V)RT`
`=(0.004284xx1000)/((250+100))xx0.0821xx220`
`=0.221 atm`