Question

A 0.1 m3 spherical tank containing gas A is initially at 0.05 MPa and 400 K. The chemical reaction A  2R is initiated. Species R is also gaseous. After species A is 50% reacted, the temperature has fallen to 350 K. What is the pressure in the vessel? Use the ideal gas law.

Answer 1

1atm =0.101325 Mpa

1Mpa= 1/0.101325 atm=9.869 atm

0.05Mpa= 0.05*9.869=0.49 atm

T= 400K and V=0.1m3=100L

From PV= nRT where, n= number of moles and R=0.08206 Lit.atm/mole.K

n= PV/RT= 0.49*100/(0.08206*400)=1.493moles. These are the mole of A initially present

when 50%A has reacted, 2Mole of R are formed, R formed = 2*1.493*0.5=1.493 A remaining= 1.493*0.5=0.7465

Total moles after 50% of the reaction =mole of R formed + Moles of A = 1.493+0.7465=2.2395

given T= 350K after 50% conversion, V=100L

P= nRT/V= 2.2395*0.08206*350/100=0.64 atm=0.0648 Mpa