An ideal gas occupies an unknown volume V liters (L) at a pressure of 12 atm. The gas is expanded isothermally against a constant external pressure of 2 atm so that its final volume becomes 3 L. The work involved for this expansion process is ....cal. (Round off to two decimal places) Related: General Organic Chemistry - General Organic Chemistry, Organic Chemistry?

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An ideal gas occupies an unknown volume V liters (L) at a pressure of 12 atm. The gas is expanded isothermally against a constant external pressure of 2 atm so that its final volume becomes 3 L. The work involved for this expansion process is ....cal. (Round off to two decimal places)

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An ideal gas occupies an unknown volume V liters (L) at a pressure of ...

Solution:

Given data:
Initial pressure, P1 = 12 atm
Initial volume, V1 = ?
Final volume, V2 = 3 L
External pressure, Pext = 2 atm

Isothermal process:
For an isothermal process, the temperature remains constant. Therefore, the ideal gas equation is given as:
PV = nRT
Where,
P = pressure
V = volume
n = number of moles
R = universal gas constant
T = temperature

Since the temperature remains constant, we can write:
P1V1 = P2V2
Where,
P1 = initial pressure
V1 = initial volume
P2 = final pressure
V2 = final volume

Solving the above equation for V1, we get:
V1 = P2V2/P1
V1 = (2 atm)(3 L)/(12 atm)
V1 = 0.5 L

Work done:
The work done for an isothermal expansion is given by:
W = -nRT ln(V2/V1)
Where,
n = number of moles
R = universal gas constant
T = temperature
V1 = initial volume
V2 = final volume

Since the temperature remains constant, we can write:
W = -nRT ln(V2/V1)
W = -(Pext)(V2 - V1) ln(V2/V1)
W = -(2 atm)(3 L - 0.5 L) ln(3 L/0.5 L)
W = -10.77 cal

Therefore, the work involved for this expansion process is -10.77 cal (negative sign indicates work done by the system).

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