Question

10 moles of an ideal gas at 27°C and 10atm pressure occupying a volume of 24.6 L. Undergoes the following changes: 1) ISOTHERMAL and reversible expansion to 246L 2) ISOTHERMAL AND irreversible expansion to 246L 3) Isochoric heating to 177C Calculate the work done in each transfirmation in KJ

Answer 1

GIVEN:

• Moles= n= 10
• Temperature= 27°C =300K
• Initial Volume= 24.6L

• R= Gas constant = 8.31

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Solution:

1)

Work done in isothermal and reversible expansion:

=>

$W = - 2.303 \times n \: R \: T \: log \dfrac{V2}{V1}$

$= > w = - 2.303 \times \: 10 \times 8.31 \times 300 \times log \: \dfrac{246}{24.6}$

by simplifying we get,

$= > - 57413.79 \: j$

$\bold{\red{= > - 57.41kJ}}$

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2)

Work done in isothermal irreversible expansion:

W = -P(V2-V1)

=> W= -1×(246-24.6) L-atm

=> W= -221.4 L-atm

=> W= -221.4 × 101.3 J

=> W= $\bold{\red{-22.43 kJ}}$

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3)

we know in Isochoric process:

• Change in Volume = 0

Work done in Isochoric process is:

= P × Change in volume

= P × 0

$\bold{\red{= 0 kJ}}$