Question

Predict the sign of ΔS for each of the following systems, which occur at constant temperature I. The volume of 2.0 moles of Ne (g) increases from 54 L to 62 L. II. The pressure of 2.0 moles of Ne (g) increases from 0.90 atm to 1.0 atm.

a) I: ΔS = negative; II: ΔS = negative

b) I: ΔS = positive; II: ΔS = negative

c) I: ΔS = positive; II: ΔS = positive

d) I: ΔS = negative; II: ΔS = positive

Answer 1

To solve this problem we will use the following ΔS relation ,

The above equation is the most common equation used to calculate entropy change for an ideal gas,

Now since the temperature is not changing,

Ln(T2/T1) =LN(1) =0

Hence the ΔS equation will reduce to

Now let u recal ideal gas law,

P V = N R T

Where , P = pressure, V= volume, N = number of moles, R is ideal gas constant , T = Temperature

Since temperature is constant , For two points (1) and (2) we can write

P1V1=P2V2

Hence, (P2/P1) =(V1/V2)

Hence ΔS equation can also be written in terms of volume as

(I) The volume of 2.0 moles of Ne (g) increases from 54 L to 62 L

This means that V1= 54 , V2 =62

Hence (V1/V2 )= (54/62)=0.87096

Therefore , ΔS =-R*ln(0.87096) =+0.138*R

Now R is a ideal gas constant which is a postive constant, Hence ΔS = positive '+'

(I) ΔS Positive

(II) The pressure of 2.0 moles of Ne (g) increases from 0.90 atm to 1.0 atm.

Ler us recall ΔS in terms of pressure

now, P1= 0.9 atm , P2=1 atm

Hence, ΔS = - R *ln(1/0.9) = - R*Ln(1.11) = -0.10536*R

R is an ideal gas constant which is a positive constant, hence ΔS= negative ,

(ii) ΔS Negative

Hence , correct answer is option (b) (I) ΔS Positive (ii) ΔS Negative