Acylinder with a piston contains 0.250 mol of oxygen at2.40x105 pa and 355 k. the oxygen may be treated as an ideal gas. the gas first expands isobarically to twice its original volume. itis then compressed isothermally back to its original volume, andfinally it is cooled isochorically to its original pressure.

(a)show the series of processes on a pv diagram.
(b) compute thetemperature during the isothermal compression.
(c) compute themaximum pressure.
(d) compute the total work done by the piston onthe gas during the series of processes.

= 285 Joules

Explanation:

a) answer can be found out in attachment

(b) The temperature for the isothermal compression is the same as the temp at the end of the isobaric expansion. Since pressure is held constant but volume doubles, we use the ideal gas law:

p V = nR T          to see that the temperature also doubles.

.So...   temp for isothermal compression =   355×2 = 710 K

.(c)   The max pressure occurs at the top point. At this point, the volume is back to the original value but the temperature is twice the original value. So the pressure at this point is twice the original, or

max pressure = 2×240000 Pa =  480000 Pa  =   4.80 x 10^5 Pa

(d) total work done by the piston = workdone during isothermal compression - work done during expansion =

=  nRT ln(V initial / V final)-p (V initial - V final)

=   nRT ln(2) - nR(T final - T initial)  

= 0.250× 8.314 ×710×ln(2)-0.250×8.314× (710 - 355)

=     285 Joules

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