When the first law of thermodynamics is applied to an ideal gasthat is taken through an isochoric process,
a. w = 0
b. delta u = 0
c. q = 0
d. none of the above

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Patrina SchowalterLv2

14 Apr 2019

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A heat engine consists of two isothermal paths (ab withT_H = 1000 K and cd with T_L = 300 K)and two isochoric paths (bc and da). Assume thatthe working substance is 2 moles of an ideal monatomic gas. Thework done by an ideal gas along sothermal paths can be calculatedusing the formula W = nRT ln(V_final/V_initial) where T is the constanttemperature along the isotherm and V_final(V_initial) is the volume of the gas in the final(initial) state. Using the first law of thermodynamics, Eq. (??),and the formula for heat during constant volume processes, (Q =nC_v ÎT) determine the following:

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[Helpful Hints: Use the First Law of Thermodynamics: Delta U = Q -W]
Helpful Data and Equations:
Q = mL LV = 22.6*105 J/kg W = pDeltaV 1 L = 10-3 m3

The first law of thermodynamics is Q = ∆U + W. For each of the following cases, state whether the internal energy of an ideal gas increases, decreases, or remains constant: (a) No energy is transferred to the gas as it expands to twice its original volume. (b) The gas volume is held constant while energy Q is removed. (c) The gas volume is held constant and no energy is transferred to or from the gas. (d) The gas temperature increases.