a mass of m = .24kg of steam is contained in a closed cylindrical container is c

Question

a mass of m = .24kg of steam is contained in a closed cylindrical container is compressed by a piston from initial temperature of T_1=280 degrees C and p_1=1 bar to a final temperature of T_2=400 degrees C. Assume that the steam is an ideal gas in answering the following questions.
a) If the pressure and specific volume are assumed to be related by a polytropic expression of the form pv^n = const for this process, where n = 1.4, determine the final pressure p_2, and the work done during the process.
b) Determine the amount of heat transfer Q that must be done during this proccess for the given initial and final conditions.

Explanation / Answer

a) here pv^1.4 = constant

since it is an ideal gas so pV = nRT

so, p^(1-n)T^n = constant

so, (p1/p2)^(1-n) = (T2/T1)^n

or, (1/p2)^-0.4 = (673/553)^1.4

or, p2 = 1.82 bar

work done = integral of Pdv, solving you get:

W = (p1v1-p2v2)/(n-1) = R(T1-T2)/(n-1) = 8.314*(673-553)/0.4 = 2494.2 J

b) du = dq - dw

du = Cv(T2-T1) = R(T2-T1)/(-1) , where =Cp/Cv for given gas

so dq = du + dw = R(T1-T2)[1/(n-1) - 1/(-1)]

For steam = 1.28

so Q = 8.314*(673-553)(1/0.4 - 1/0.28) = -1068.94 J

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