Question

of water at 290k. Calculate final temp
constant
A Gas Expands from 31 to 50 against
of 3 atm the work done
heat 10 mole
pressure
during Expansion is used to heat lomde
. ا-را-
of water. s=4.18 J
S =4.1859-1kY​

Answer 1

Explanation:

Work done =P×dV=3.0×(5.0−3.0)

=6.0litre−atm=6.0×101.3J

=607.8J

Let ΔT be the change in temperature

Heat absorbed =m×s×ΔT

=10.0×18×4.184×ΔT

Given, P×dV=m×s×ΔT

or ΔT=

m×s

P×dV

=

10.0×18.0×4.184

607.8

=0.807

Final temperature=290+0.807=290.807K

hope you understand it bro do hardwrok all the best