Determine the efficiency for the cycle shown in the figure, using the definition. Assume that the gas in the cycle is ideal monatomic gas.
Compare with the efficiency of a Carnot engine operating between the same temperature extremes.
the cycle is composed of two isobaric and two isovolumetric paths.
work, W = 0
heat, Q = deltaU = nC_v(deltaT) where deltaU is change in internal energy, n is number of moles, C_v is constant volume (monotomic so = 3R/2 where R = 8.314), delta T is change in temp
work, W = p(V_2 - V_1) where p is pressure in pascals, V_2 is final volume, V_1 is initial volume
heat, Q = deltaU + W = nC_p(deltaT) where C_p is constant pressure = C_v + R
ideal gas law: PV = nRT
efficiency, e = W/Q_h = (Q_h - Q_c)/Q_h = 1 - (Q_c/Q_h) where W is work, Q_h is heat energy (high), and Q_c is cool heat energy (low)
The Attempt at a Solution
i intend to use efficiency, e = W/Q_h, since W = 0 for each of the two isovolumetric paths, the work only comes from the two isobaric paths, i calculated:
top horizontal path to be W = 2431.8 joules, and the bottom horizontal path to be W = 1215.9 joules, so net work would be W_net = 3647.7 joules
now in order to solve for Q_h or even Q_c for that matter I need to be able to solve for heat Q, in order for that i need to know the number of moles and the temp in kelvin.
since i am only concerned with Q_h, do i only need to know the heat energy Q from the top horizontal (isobaric) path?
so Q = nC_p(deltaT) and using ideal gas law and solving for T and subsituting in for T is the constant pressure heat equation i get:
Q = nC_p(PV/nR) so the n's cancel out, and C_P = C_v + R = 3(8.314)/2 + 8.314 = 20.785, here is where i am confused. since the top horizontal path is isobaric, the volume changes thus what volume do i use for V?
so currently my equation for Q would be Q = nC_p(PV/nR) = 20.785[((607950Pa)(V))/8.314]
any help/clarification appreciated