Respuesta :
Complete Question
The complete question is shown on the first uploaded image
Answer:
The workdone is  [tex]W = -177.275J[/tex]
Explanation:
From the question we are told that
   The initial Volume is  [tex]Vi = 0.160 L[/tex]
   The final volume is  [tex]V_f = 0.510L[/tex]
   The external pressure is  [tex]P = 5.00 \ atm[/tex]
Generally the change in volume is
      [tex]\Delta V = V_f - V_i[/tex]
Substituting values we have
      [tex]\Delta V = 0.510 -0.160[/tex]
         [tex]= 0.350L[/tex]
Generally workdone is mathematically represented as
      [tex]W = -P \Delta V[/tex]
W is negative because the working is done on the environment by the system which is indicated by volume increase
   Substituting values
        [tex]W = - 5* 0.350[/tex]
          [tex]= -1.75 \ L \ \cdot atm[/tex]
Now  [tex]1 \ L \cdot atm = 101.3J[/tex]
 Therefore  [tex]W = -1.75* 101.3[/tex]
             [tex]= -177.275J[/tex]
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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!!!