A cylinder with a piston contains 0.225mol of ideal oxygen at a pressure of 2.35×10^5Pa and a temperature of 345K . The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure.
Compute the total work done by the piston on the gas during the series of processes. in J
Compute the total work done by the piston on the gas during the series of processes. in J
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work done is defined as integral pdV
work done in first process = nr(T2-T1) = 0.225*8.3*(690-345)= 644.29 J
T2 is taken as 690K because in an isobaric process, temperature is directly proportional to volume, so when volume doubles temperature also doubles.
work done in second process = nRT*lnV2/V1= 0.225*8.3*690*-0.693 = -892.98 J
as volume again becomes to previous value V2/V1=1/2 and ln 1/2 = -0.693
work done in third process = 0 J (because in isochoric process no volume change occurs and hence work done is zero)
so total work done on piston by gas = 644.29-892.98 = -248.69 J
total work done by piston on gas = - work done by gas on piston = -(=248.69) J = 248.69 J
hope it helps..
need 10 points
work done in first process = nr(T2-T1) = 0.225*8.3*(690-345)= 644.29 J
T2 is taken as 690K because in an isobaric process, temperature is directly proportional to volume, so when volume doubles temperature also doubles.
work done in second process = nRT*lnV2/V1= 0.225*8.3*690*-0.693 = -892.98 J
as volume again becomes to previous value V2/V1=1/2 and ln 1/2 = -0.693
work done in third process = 0 J (because in isochoric process no volume change occurs and hence work done is zero)
so total work done on piston by gas = 644.29-892.98 = -248.69 J
total work done by piston on gas = - work done by gas on piston = -(=248.69) J = 248.69 J
hope it helps..
need 10 points