and thereby does work on the gas. The temperature is kept constant by allowing h
ID: 1552356 • Letter: A
Question
and thereby does work on the gas. The temperature is kept constant by allowing heat transfer to and from the surroundings, and the variation of pressure as the volume changes at fixed temperature is given by the ideal gas law. The problem involves calculating the work done by a non-constant force using the ideal gas law to find the pressure at each value of the volume. The work done on the gas in an infinitesimal change of volume dV is -PdV, where P is determined for each value of the volume V and temperature by the ideal gas law: PV = nRT. The total work that the applied force does on the system is w = - integral^v_f _v_i p dv. Using the ideal gas law, we find w = - nRT integral^n_f _v_i 1/v dv = - nRT in (v_f/v_i). The constant nRT has the value P_iV-i that it had at the initial volume and temperature, leading to W = - p_i v_j in (v_f/v_i) = (1013 times 10^5 N/m^2)(1.00 times 10^-3 m^3) ln (1/5.0) = J. Notice that the work done on the system is positive. Work on the gas has to be done to compress it. The natural log dependence in the expression for the work done predicts a weaker dependence on the final and initial volume than would result from a linear relation. Compare how the area under the PV curve changes with variation in the volume V_t relative to how it would change in a linear PV diagram, noting where the specific volume dependence came from in the present results.Explanation / Answer
w = -piviln(vf/vi)
w = -(1.013*10^5*10^-3)ln(1/5)
w = -1.013*10^2ln(1/5)
w = -1.013*(-1.609)
w = 1.629 J
in linear pv diagram area is w = p*chaneg in volume
in expansion chane is volume is positive
in compression change volume is negative so w is negative
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