Find entropy change for free expantion of ideal gas

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In an isolated system, when 1 liter of ideal gas freely expands into a 2 liter volume, the energy and temperature remain constant, but the entropy increases. This apparent contradiction arises from the definition of entropy as ΔS = ∫dQ/T over a reversible path. To accurately calculate the change in entropy, one must consider a reversible process involving work and heat flow. The initial and final states of the gas must be analyzed to understand the entropy change. Thus, while the formula suggests constant entropy under certain conditions, the actual process of free expansion leads to an increase in entropy.
klinke
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entropy = jouls/kelvin
supose 1 liter of ideal gas is allowed to freely expand into a 2 liter volume in an isolated system
the energy in the system would remain the same,
the temperature in the system would remain the same
therefore if entropy =jouls/kelvin the entropy would remain the same
howerer if 1 liter of ideal gas is allowed to freely expand into a 2 liter volume in an isolated system the entropy does increase.
how dose entropy=jouls/kelvin ?
 
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klinke said:
entropy = jouls/kelvin
supose 1 liter of ideal gas is allowed to freely expand into a 2 liter volume in an isolated system
the energy in the system would remain the same,
the temperature in the system would remain the same
therefore if entropy =jouls/kelvin the entropy would remain the same
howerer if 1 liter of ideal gas is allowed to freely expand into a 2 liter volume in an isolated system the entropy does increase.
how dose entropy=jouls/kelvin ?
ΔS = ∫dS = ∫dQ/T over a reversible path between the initial and final states.

The initial state is (P,V,T) and the final state is (P/2,2V,T). So to calculate the entropy change you have to find a reversible path between those two states. (hint: the reversible path involves work being done and heat flow into the gas).

AM
 

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