1. ## urgent please.needed for tomorrow

3. The entropy change, ΔS, associated with the isothermal (i.e.
constant temperature) change in volume of n moles of an ideal
gas from volume V1 to volume V2 is given by the equation
ΔS = nR ln(V2/V1)
where R is the gas constant.
(a) If a compression occurs (V2 < V1) use Equation 1 to deduce
whether there will there be an increase (ΔS is positive) or
decrease (ΔS is negative) in the entropy of the gas.

(b) If an expansion occurs (V2 > V1) use Equation 1 to deduce
whether there will there be an increase or decrease in the
entropy of the gas.
(c) The ideal gas law states that pV = nRT, where p is the
pressure of the gas. Use this expression and Equation 1 to
obtain an equation relating the entropy change to the initial
and final pressures (p1 and p2). (Note that n,R and T will
be constant at the two different pressures).

2. hey,

for part a) entropy will decrease because if V2is smaller than V1, inside the brackets is a fraction, this will make ln(answer) a negative number and therefore the whole right side negative.

for part b) entropy will increase because if V2is larger than V1, inside the brackets is larger than 1, this will make ln(answer) a positive number and therefore the whole right side positive.

Another way to think about it logically is to think of the definition of entropy- measure of disorder or the number of ways or arranging the molecules of gas (in this case). If you compress the gas (part a) then there are less ways to arrange the molecules. If you expand, more ways to arrange.

for part c) make V the subject of the ideal gas equation: V= (nRT)/p. And since n,R and T are the same for both pressure/volume you can sub this into the change in entropy equation to get: ΔS = nR ln[(nRT)/p2/] / [(nRT)/p1] sorry for how that looks btw. The nRT cancel out leaving you with (1/p2) / (1/p1) flip (1/p1) around to make : (1/p2)*p1 which equals (p1/p2). Therefore the full equation:
ΔS = nR ln(p1/p2).