Question

2 moles of an ideal gas are expanded isothermally from a volume of 25 dm3 to 30dm3 at 300k against
a pressure of 1.2bar atm. Calculate work done in dm3 bar
and J (R = 8.314 si unit​

Answer 1

Answer:

Work done in an isothermal reversible process is given by-

W=−nRTln

V

1

V

2

Where,

n = no. of moles =1

R= molar gas constant =8.314Jmol

−1

K

−1

T= temperature =300K

V

1

=10dm

3

V

2

=20dm

3

∴W=−1×8.314×300×ln

10

20

⇒W=−2494.2×(ln2−ln1)=−1728.8 J=−1.73 kJ

Now, from first law of thermodynamics,

ΔU=ΔH+W

Since temperature is constant, i.e. ΔU=0.

∴ΔH=−W=−(−1.73)=+1.73KJ

Explanation:

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