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(A) A chemical reaction takes place in a container of cross-sectional area 100 cm2. As a result of the reaction, a piston is pushed out through 10 cm against an external pressure of 1.0 atm. Calculate the work done by the system.

(B) A chemical reaction takes place in a container of cross-sectional area 50.0 cm2. As a result of the reaction, a piston is pushed out through 15 cm against an external pressure of 121 kPa. Calculate the work done by the system.

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Answer:

(A) The work done by the system is -101.325J

(B) The workdone by the system is -90.75J

Explanation:

(A) Workdone = -PΔV

Given that A = 100cm2 = 0.01m2

distance d = 10cm = 0.1m

ΔV= Area × distance

ΔV= 0.01 ×0.1

ΔV = 0.001m3

P= external pressure = 1atm = 101325Pa

Workdone = -0.001 × 101325

W= - 101.325Pa m3

1Pam3 = 1J

Therefore W = - 101.325J

The work done on the system is -101.325J

(B) Workdone = -PΔV

Given that A = 50cm2 = 0.005m2

distance d = 15cm = 0.15m

ΔV= Area × distance

ΔV= 0.005×0.15

ΔV = 0.00075m3

P=121kPa = 121000Pa

W= - 121000 × 0.00075

W= -90.75Pa m3

1Pam3 = 1J

W = - 90.75J

The woekdone by the system is -90.75J

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