A 2.0-mol sample of ideal gas with molar specific heat cv= 5/2R is initially at

Question

A 2.0-mol sample of ideal gas with molar specific heat cv= 5/2R is initially at 300K and 100kPa pressure. Okay this was a multi part question: I am only stuck on the one stared w/o an 1st part : Determine the final temperature when 1.3 of heat are added to the gas isobarically.=320K (this is correct) I did q=ncpdeltat deltat=1300J/58.198J=22.3 + 300 = 320J... this is correct f)Determine the work done by the gas when 1.3 of heat are added to the gas isobarically.????HELP i did work= p(v2-vi) and got 5416995012.... this is not correct, help??

Explanation / Answer

A) Heat added Isothermally: in this case, the change in internal energy is zero and from the 1st law, the heat added equals the work produced and, thus, W = Q = 1.3 B) Heat added at constant volume: in this case, there is no work in or out and, thus, the change in internal energy equals the heat input, this is U = cv (T2 - T1) = Q solving for T2 we get T2 = Q/cv + T1 you have all the data now, substitute and find T2 C) Work = 0 when process is at constant volume D) Heat added isobarically: From the 1st law we have U = Q - W = cv (T2 - T1) = Q - P(V2 - V1) Equation 1 is then (1) cv (T2 - T1) = Q - P(V2 - V1) since this is an isobaric process, from the general equation of ideal gases we have V2 / T2 = V1 / T1 = n R T1 / P Equation 2 is then (2) V2 / T2 = R T1 / P The unknowns are V2, and T2, all the other quantities are known, use thus eqs (1) and (2) to obtain V2 and T2. E) Work = P (V2 - V1) V1 = n R T1/P1

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