Question
Question: The entropy change when 1 mole of ideal gas at 27\(^o\)C is expanded reversibly from 2 litre to 20 l...
The entropy change when 1 mole of ideal gas at 27oC is expanded reversibly from 2 litre to 20 litre is
(A) 2calK−1mol−1
(B) 4.6calK−1mol−1
(C) 4JK−1mol−1
(D) 4.6kJK−1mol−1
Solution
Hint : Entropy of the system is defined as the randomness or degree of freedom of any system. Entropy is denoted by ‘S’. Entropy is a thermodynamic quantity which depends on thermodynamic variables that are temperature, pressure and volume. We cannot calculate the absolute value of entropy for any system but we can calculate the change in entropy
( ΔS ).
Complete Step By Step Answer:
Thermodynamically change in entropy is represented as
ΔS=Tqreversible . This is the formula to calculate change in entropy.
Where,
qreversible = heat exchange for the system which is reversible in nature (Joule or calorie).
T = temperature(K).
The unit of entropy is JK−1 or CalK−1 .
1JK−1 is said to be 1 entropy unit.
Given: ΔS =?
n=1mol
T= 270C = 27+273=300K .
V1=2L
V2=20L
Let’s consider R=2calmol−1K−1
Let’s rewrite the equation,
ΔS=Tqreversible
Substituting q by ΔU+PΔV ,
ΔS=TΔU+PΔV
For an ideal gas ΔT=0,ΔU=0 .
So, we get
ΔS=TPΔV
From the ideal gas equation we know that P=VnrT
ΔS=V×TnRT×ΔV
ΔS=nRVΔV
On integration we have,
ΔS=nRlnV1V2 .
Substituting the given values we have,
⇒ΔS=1×2×ln220
ΔS=2×2.303
ΔS=4.606calK−1mol−1 .
So, the correct option is (B).
Note :
For any reversible system there is one and only path. So we can say that qreversible is a state function. Change in entropy is inversely proportional to temperature that is at low