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ruslelena [56]
2 years ago
13

The joule (J) is a unit of energy. Recall that energy may be converted between many different forms such as mechanical energy, t

hermal energy (heat), chemical energy, electrical energy, and light.The mechanical energy produced by a system is called work. When work is accomplished through the changing volume of a gas, it is called PV work and is given by the formula w=−PΔV, where w is the work, P is the external pressure, and ΔV is the change in volume. If the volume change and pressure are in liters and atmospheres respectively, then the work will have units of liter-atmospheres, which can be converted to joules using the conversion factor 1 L⋅atm=101.3 J.Work can also be expressed as force multiplied by distance: w=F×d, where w is the work in joules, F is the force in newtons, and d is the distance in meters. Note that 1 N⋅m=1 J.

Physics
2 answers:
REY [17]2 years ago
7 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The workdone is  W = -177.275J

Explanation:

From the question we are told that

      The initial Volume is  Vi = 0.160 L

      The final volume is  V_f = 0.510L

      The external pressure is  P = 5.00 \ atm

Generally the change in volume is

           \Delta V = V_f - V_i

Substituting values we have

           \Delta V = 0.510 -0.160

                 = 0.350L

Generally workdone is mathematically represented as

           W = -P \Delta V

W is negative because the working is done on the environment by the system which is indicated by volume increase

     Substituting values

                W = - 5* 0.350

                    = -1.75 \ L \ \cdot atm

Now  1 \  L \cdot atm = 101.3J

  Therefore  W = -1.75* 101.3

                          = -177.275J

   

svetoff [14.1K]2 years ago
5 0

Answer:

Workdone = -1.75 L.atm = -177.275 J

Explanation:

Given,

External pressure = 5.00 atm

Initial volume = V(initial) = 0.160 L

Final volume = V(final) = 0.510 L

The piston acts against the constant external pressure of 5.00 atmospheres. The volume expands to 0.510 L from 0.160 L

W = -PΔV

P = 5.00 atm

ΔV = V(final) - V(initial) = 0.510 - 0.160 = 0.35 L

W = - 5 × 0.35 = - 1.75 L.atm

This work can then be expressed in Joules

1 L.atm = 101.3 J

-1.75 L.atm = -1.75 × 101.3 = -177.275 J

Note that the workdone is negative because the system does work on the environment in this expansion procase that led to increase in volume.

Hope this Helps!!!

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