Answer
(a) $0.51\times 10^3J$
(b) Same
(c) $-0.51\times 10^3J$
Work Step by Step
(a) We can find the required work done as follows:
$W=P(V_f-V_i)$
We plug in the known values to obtain:
$W=(160\times 10^3Pa)[(8.60\times 10^{-3}m^3)-(5.40\times 10^{-3}m^3)]$
$W=0.51\times 10^3J$
(b) We know that the work done is directly proportional to the change in volume, if the pressure is kept constant. Since it is given that the volume remains the same, the work done will be the same as found in part(a).
(c) We can find the required work done as follows:
$W=P(V_f-V_i)$
We plug in the known values to obtain:
$W=(160\times 10^3Pa)[(5.40\times 10^{-3}m^3)-(2.20\times 10^{-3}m^3)]$
$W=0.51\times 10^3J$
