where we have used the ideal gas law in the last step. We emphasize that, in order to obtain work and energy in Joules, pressure should be in Pascals (N / m2) and volume should be in cubic meters. The degrees of freedom for a diatomic gas is f= 5. (a) The internal energy change is
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3333int int 3552.010 Pa4.0m5.010 Pa2.0m225.010J.caccaaEEp Vp V (b) The work done during the process represented by the diagonal path is 33diag= 3.5 10 Pa2.0m2accappWVVwhich yields Wdiag= 7.0×103J. Consequently, the first law of thermodynamics gives 333diag intdiag(5.0107.010) J2.0 10J.QEW (c) The fact that Eintonly depends on the initial and final states, and not on the details of the ―path‖ between them, means we can write 3intint int 5.0 10JcaEEE for the indirect path, too. In this case, the work done consists of that done during the constant pressure part (the horizontal line in the graph) plus that done during the constant volume part (the vertical line): 334indirect5.0 10 Pa2.0m01.0 10J.WNow, the first law of thermodynamics leads to 343indirectintindirect(5.0101.010) J5.0 10 J.QEW 11. Suppose the gas expands from volume Vito volume Vfduring the isothermal portion of the process. The work it does is ifVVVVVVnRTVdVnRTpdVWfifiln, where the ideal gas law pV = nRTwas used to replace pwith nRT/V. Now Vi= nRT/piand Vf= nRT/pf, so Vf/Vi= pi/pf. Also replace nRTwith piVito obtain ln.iiifpWpVpSince the initial gauge pressure is 1.03 105Pa, pi= 1.03 105Pa + 1.013 105Pa = 2.04 105Pa. The final pressure is atmospheric pressure: pf= 1.013 105Pa. Thus 553452.0410 Pa2.0410 Pa0.14mln2.0010 J.1.01310 PaWDuring the constant pressure portion of the process the work done by the gas is W= pf(Vi–Vf). The gas starts in a state with pressure pf, so this is the pressure throughout this portion of the process. We also note that the volume decreases from Vfto Vi. Now Vf= piVi/pf, so .1044.1)14.0)(1004.210013.1()(4355JmPaPaVpppVpVpWiiffii
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