Apiston has an external pressure of 15.0 atm. how much work has been done in joules if the cylinder goes from a volume of 0.150 liters to 0.640 liters?

Work, in thermodynamics, is the amount of energy that is transferred from one system to another system without transfer of entropy. It is equal to the external pressure multiplied by the change in volume of the system. It is expressed as follows:

W = PdV

Integrating and assuming that P is not affected by changes in V or it is constant, then we will have:

W = P (V2 - V1)
W = 15 (101325) N/m^2 ( 0.640 - 0.150 ) (1/1000) m^3
W = 744.74 N-m or 744.74 J

It is important to remember that pressure here is assumed to be independent of the volume. However, if this is not the case the formula of work would change accordingly.

Therefore, the work done on the cylinder is equal to 744.74 J.

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