Question

If pressure of 3 mol of gas is reduced from 1 atm to 0.5 atm
at constant temperature of 25°C, the change in entropy is:​

Answer 1

Answer:

sorry I don't know how to do this one

Explanation:

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Answer 2

Answer:

The change in entropy is 28.8 J/K.

Explanation:

Step-1: Change in entropy, ΔS = $\frac{5}{2}$ nR ㏑$\frac{T2}{T1}$ - nR In$\frac{P2}{P1}$ ..................(1)

• n = number of moles = 3 moles
• R =universal gas constant= 8.31 J$K^-1$$mol^-1$
• Temperature, T1 =T2 = constant
• P1 = initial pressure = 1 atm
• P2 = final pressure = 0.5 atm

Step-2:     $\frac{T1}{T2}$ = 1 and ㏑1 = 0

               $\frac{P2}{P1}$ = $\frac{0.5}{1}$ = $\frac{1}{2}$  and ㏑$\frac{1}{2}$ = -0.693

Step-3:   Putting all the values in eqt (1),  

                ΔS = ( $\frac{5}{2}$ nR X 0) - (nR X -0,693)

                      = 0 + (3 X 8.31 X 0.693)

                      = 28.79 = 28.8 J/K

Change in entropy is 28.8 J/K.