Question

an ideal gas expands isothermally from 1L to 10L at 25 degree C . find the value of change in enthalpy???
IRRELEVANT ANSWERS WILL BE REPORTED!!!

Answer 1

Answer:

Work done in a reversible isothermal process is:

W=−2.303nRTlog

V iV f .....(1)

Given:-

n=2 moles

T=300K

V f =10L

V i =1L

R= Gas constant =8.314J/K−mol

Substituting these values in eq n

(1), we have

W=−2.303×2×8.314×300×log 1 10

⇒W=−11488.285J=−11.4kJ

Now as we know that,

ΔH=ΔU+W

For an isothermal process,

ΔU=0

∴ΔH=W=−11.4kJ

Hence the enthalpy change for the given process is −11.4kJ.

Hope it helps u ✌︎

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