Question

A gas absorbs 500J heat and utilized QJ in
doing work against an external pressure of 2
atm. If ∆E is -510 J, values of ∆V and W
respectively are​

Answer 1

Answer:

Explanation:

work done by reference is taken positive

and in chemistry it is external agent

so work done by external is taken positive

and work done against external agent is taken negative

so work done = -q j

by law of thermodynamics

delh=delu+delpv

delh=delu+pdelv+vdelp

delh=delu+pdelv

500=-510+pdelv

1010 = pdelv

and we know that lt atm = 100j

and 1 joule = 0.01 lt atm

pdelv=0.01*1010

pdelv=10.1

delv=5.05 lt

and work is against external agent so negative work

w=-pdelv

w=-10.1 lt atm