Answer:
The partial pressure of argon in the flask = 71.326 K pa
Explanation:
Volume off the flask = 0.001
Mass of the gas = 1.15 gm = 0.00115 kg
Temperature = 25 ° c = 298 K
Gas constant for Argon R = 208.13
From ideal gas equation P V = m RT
⇒ P =
Put all the values in above formula we get
⇒ P = × 208.13 × 298
⇒ P = 71.326 K pa
Therefore, the partial pressure of argon in the flask = 71.326 K pa
Explanation:
Assuming that moles of nitrogen present are 0.227 and moles of hydrogen are 0.681. And, initially there are 0.908 moles of gas particles.
This means that, for
moles of + moles of
= 0.908 mol
Since, 2 moles of =
= 0.454 mol
As it is known that the ideal gas equation is PV = nRT
And, as the temperature and volume were kept constant, so we can write
=
=
=
= 5.2 atm
Therefore, we can conclude that the expected pressure after the reaction was completed is 5.2 atm.