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How do you calculate ΔSsys for the isothermal expansion of 1.5 mol of an ideal gas from 20.0 L to 22.5 L?

Answer
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Hint: The measure of disorder or randomness in a system is known as entropy of the system. Properties of a thermodynamic system such as pressure, heat capacity, and temperature can be described using entropy. Its SI unit is J/mol.

Complete answer:
When we divide the amount of heat that is absorbed or released in an isothermal and reversible process by the absolute temperature, we get the value of entropy change.
ΔS=QT
Now, we know that an ideal gas is a gas in which
- the molecules either attract or repel each other
- the molecules do not occupy any space themselves.
So, when the volume of an ideal gas increases without any changes in temperature, it is known as the isothermal expansion of an ideal gas.
Heat change during the isothermal expansion of an ideal gas is given by
Q=nRTVV
This equation can be re-written as
QT=nRVV
Also, we know that
ΔS=QT
So, by using these equations, we can say that the change in entropy when gas is expanded isothermally and increases in volumes is given by
ΔS=nRV1V2VVΔS=nRlnV2V1
Where ΔS is the change in entropy, n is the number of moles, R is the gas constant, V2 is the volume of gas after expansion, and V1 is the volume of gas before expansion.
It is given to us that
n = 1.5 moles,
V1= 20 L,
V2 = 22.5 L,
And R = 6.314 J/mol K
So, by using the formula, we get
ΔS=1.5×8.314×ln22.520J/KΔS=1.47 J/K
So, the change in entropy ΔSsys for the isothermal expansion of 1.5 mol of an ideal gas from 20.0 L to 22.5 L is 1.47 J/K.

Note:
It should be noted that when we add the entropy changes of the system and surroundings, we get the total entropy change. It is given by
ΔStotal=ΔSsys+ΔSsur
When the total entropy change of a reaction is positive, it indicates that the process is spontaneous.
When the total entropy change of a reaction is negative, it indicates that the process is nonspontaneous.
When the total entropy change of a reaction is 0, it indicates that the process is at equilibrium.
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