Hope, someone can help. Hey. I am to show that the change in entropy for a mono

Question

Hope, someone can help.

Hey. I am to show that the change in entropy for a mono atomic ideal gas, when expanding quasistatically and isothermally is: Delta S = Nkb ln(Vf/Vi) . Hope, someone can help. W = -P delta V , and the formula volume dependent change in entropy, where U and N are held fixed: Delta U = Q + W (not dQ and dW! My teacher hates that notation!), the formula for expansion work PV = NkbT , the first law of thermodynamics Delta S = Q/T . by means of the following equations: The ideal gas law:

Explanation / Answer

From the first law, you should know that:

dU = ?q + ?w

where dU is the infinitesimal change in energy due to an infinitesimal amount of work (?w) done on the system and and infinitesimal amount of heat (?q) added to the system.

You should also know that the internal energy of an ideal gas depends only on temperature, so in an isothermal process, the internal energy remains constant, and we have:

dU = 0 = ?q + ?w

?q = -?w

The work done on the system is equal to -PdV, so:

?q = PdV

You appear to know that for a reversible process,

dS = ?q/T, so, for this isothermal process involving an ideal gas, we have that

dS = (P/T)dV

If we could integrate this, we'd have an expression for ?S in terms of ?V, but to do the integral, we need to know how P varies as a function of V. For this, we can use the ideal gas law:

P = n*R*T/V

so:

dS = (n*R/V) dV

Integrate to get:

?S = n*R*ln(V_final/V_initial)

This gives us an equation for the change in entropy for an isothemal change of n moles of an ideal gas from a volume of V_initial to a volume of V_final. Plug in the numbers appropriate for this case, and you have your answer.
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