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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Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge