Carnot cycle, heat and monatomic ideal gas

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A monatomic ideal gas is utilized in the Carnot cycle, with isothermal processes A to B and C to D, and adiabatic processes B to C and D to A. The work done by the gas during the isothermal process A to B is 400 J, leading to the inquiry about the heat expelled during process C to D. The efficiency of the Carnot cycle is defined as 1 - (Ql/Qh) and 1 - (Tl/Th), indicating that the heat expelled must be less than 400 J. The final calculation determined that 100 J of heat is expelled during the process C to D, with an alternative solution found using the adiabatic equation. The discussion highlights the application of thermodynamic principles in solving the problem.
frznfire219
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Hi, I would appreciate any help with this:

A monatomic ideal gas is used as the working substance for
the Carnot cycle. Processes A => B and C => D
are isothermal, while processes B => C and D => A are adiabatic.
During process A => B, there are 400 J of work done by the gas on
the surroundings. How much heat is expelled by the gas during process C => D?

So I'm completely stuck, all I know is that it's less than 400 J, obviously.
There's a picture of the corresponding PV graph actually at http://www.compadre.org/psrc/evals/Physics_Bowl_2003.pdf (page 12).

Thanks for any help!
 
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I'm not sure about this explanation, but whatever. You know that the carnot cycle runs at perfect efficiency ie. 1-(Ql/Qh) and you know that efficiency of the carnot cycle is also 1-(Tl/Th). I get 100 J.
 
Thanks! I actually figured it out later with the adiabatic equation (PV^gamma is constant) but your way is much more elegant.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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