Question

A biochemical reaction triggers the expansion of 2.5 mol of an ideal gas reversibly,coming to rest at an equalized pressure of 58 kPa at 37 C. If the initial volume was 55.5 litres,then the work that was done by the system in this process was approximately
(A) - 4466 J (B)-3600 J

Answer 1

Given:

1. Initial Volume (V₁) = 55.5

2. n = 2.5 mol

3. T₁ = T₂ = 37° C = 310 K

4. Final Pressure (P₂) = 58 kPa    

To find:

The work done by the system.            

Solution:

• We know that, PV = nRT

        P₁V₁​=nRT₁​

         P₁ ​ =​ nRT₁ / V₁​​

              = 116.09

• Now, work done = ∫ P dv

                                    = ∫ nRT / V dv

                                    = nRT ln ( V₂ / V₁ )

                                    = nRT ln ( P₁ / P₂ )

                                    = 2.5 x 8.314 x 310 ln (116.09 / 38 x 10³)

                                    = − 40030 J

• Therefore the work done by the system will be - 40030 J.

Answer 2

Answer:

Solution - Step - 1 - write the initial conditions given Initial moles = 2.5 ; Pressure final = 58 * 1000 Pa = 58KPa ; As for the entire process the temperature occurred at 37 degree celsius and hence the process is isothermal Step -2 calculation of initial pressure Initial volume(V1) = 55.5 litre = 55.5 * 10 (^-3) m^3 ; Initial Temperature = 37 degree celsius = 310 K ; Initial pressure (P1) = nRT/V1 = (2.5) R (310) / (55.5) (10 (^-3)) ; As volume is in litre so use R = 8.314 JK-1mol-1 ; Here we are actually taking all the values to be in SI unit So P1 = (2.5 *8.314 *310 ) / (55.5 * 10^-3) Pa ; P1 = 116.1 kPa ; Step -2 calculation of work done So work done = - 2.303nRT log(P1 / P2 ) ; W = -2.303 * 8.314 * 2.5* 310 *log (116.1 / 58) ; W = -4466 Joule Option-A

Explanation:

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