Question

A Joule expansion refers to the expansion of a gas from volume
*V*_{1} to volume *V*_{2} against no
applied pressure, and is sometimes also called a free expansion.
There is no work done, because the *P* of -*PdV* is
zero. By insulating the system, this process can be done
adiabatically, so there is no change in heat. For an ideal gas, the
adiabatic process is also isothermal, so there is no change in
thermodynamic energy, ∆*U* = 0 (which is not surprising,
since both *q* and *w* are 0). Because entropy is a
state variable, the change of the system's entropy can be
calculated using the most convenient means possible, which would be
through a reversible process, and is just given by the usual
expression for a reversible isothermal expansion, ∆*S* =
*nR* ln(*V*_{2}/*V*_{1}).
Thus, even though there is no heat flow, there is certainly an
increase in entropy of the system.

Explain how this overall result is consistent with the 2nd Law of Thermodynamics. Is this process reversible or irreversible? Explain your reasoning.

Answer #1

As per the 2nd Law of Thermodynamics, in a cyclic process , the entropy will either increase or will remain the same

In a free expansion there is certainly an increase in volume of the gas and thus , the randomness of the gas molecules increases and so does the entropy.

Similarly, in isothermal/adiabatic expansion, altough ∆E = 0, there is increase in volume of the gas due to work done and thus, the randomness of the gas particles increases and so does the entropy.

Thermodynamics Question
A Joule expansion refers to the expansion of a gas from volume
V1 to volume V2against no
applied pressure, and is sometimes also called a free expansion.
There is no work done, because the P of -PdV is
zero. By insulating the system, this process can be done
adiabatically, so there is no change in heat. For an ideal gas, the
adiabatic process is also isothermal, so there is no change in
thermodynamic energy, ∆U = 0 (which...

Suppose 4.00 mol of an ideal gas undergoes a reversible
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Calculate the total change of entropy for an ideal monatomic gas
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A student wrote:
If I expand an ideal gas from V1 to V2 under constant external
pressure and pex, then the work done
is w = -pex(V2 - V1). As a result, if the expansion
is isothermal, then as delta U = q + w = 0, then
q = -w = pex(V2 - V1). Therefore, delta S
=qrev/T=(p/T)(V2 - V1).
Is this statement correct or incorrect? Explain your answer.

An ideal gas at 300 K has a volume of 15 L at a pressure of 15
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(4) the change in internal energy when the gas undergoes
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b.- A reversible adiabatic expansion to a pressure of 10
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d. The internal energy of a system is only affected by heat and
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