Question

Plss answer this guys.......


A vessel of uniform cross-section of length
500 cm as shown in figure is divided in two
parts by a weightless and frictionless piston
one part contains 5 moles of He(g) and other
part 2 moles of H2(g) and 4 mole of O2(g)
added at same temperature & pressure in
which reaction takes place finally vessel
cooled to 300 K and 1 atm. What is the length
of N2 compartment?
(Assume volume of piston and vol. of H2O(1)
formed are negligible)
He
Hand
(A) 187.5
(C)312.5 cm
(B) 300 atm
(D) None of these​

Answer 1

Answer:

312.5 cm

Explanation:

Given A vessel of uniform cross-section of length

500 cm as shown in figure is divided in two

parts by a weightless and frictionless piston

one part contains 5 moles of He(g) and other

part 2 moles of H2(g) and 4 mole of O2(g)

added at same temperature & pressure in

which reaction takes place finally vessel

cooled to 300 K and 1 atm. What is the length

of He compartment?

We know that moles of oxygen remaining after reaction will be 4 – 1 = 3

V  n This is due to same temperature and pressure in each compartment after reaction.        

So length of He compartment will be

5 / 5 + 3 x 500  

= 5/8 x 500

= 312.5 cm