Homework Help: Reversable process with adiabatic expansion

1. Sep 19, 2011

osheari1

1. The problem statement, all variables and given/known data
One mole of a monatomic ideal gas is taken through the reversible cycle shown below.
http://img508.imageshack.us/img508/2595/newbitmapimageqm.png [Broken]
process bc is an adiabatic expansion, with P_b = 10.00 atm and V_b = 10^-3 m^3.
Find (a) the energy added to the gas as heat. (b) find the energy leaving the gas as heat, (c) the net work done by the gas, and (d) the efficiency of the cycle.

2. Relevant equations
PV=nRT (ideal gas)
PV^(gamma) = constant
gamma = C_p/C_v
V^(gamma-1)T = constant
U = (3/2)nRT (for a monatomic gas)

3. The attempt at a solution
I really dont know how to start this problem
I know for the bc section of the cycle Q = 0 because its an adiabatic expansion.
I also solved for the temperature at point b T = PV/(nR) = (1.0132*10^6Pa)(10^-3m^3)/(1mole)(8.315) = 121.85 K

from here im not sure what to do because I am not given gamma or the pressure at a or c

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 5, 2017
2. Sep 19, 2011

Andrew Mason

[Note: A reversible cycle has to be in equilibrium with its surroundings at all time. This means heat flow can only occur when there is an infinitessimal temperature difference between the system and the hot or cold reservoirs. So heatflow has to occur isothermally. This cannot occur on the isochoric or isobaric parts of this cycle so I fail to see how this can be a reversible cycle. But that is just an aside.]

You have to determine the temperatures at a, b and c.

First, determine the temperature at b using the ideal gas law.

Second, determine the temperature and pressure at c using the adiabatic condition.

Then determine the temperature at a from that.

Since a-b is isochoric you can work out the heatflow into the gas from the change in temperature (using what specific heat capacity and what value?)

Since c-a is isobaric, you can work out the heatflow out of the gas from the change in temperature (using what specific heat capacity and what value?).

Since b-c is adiabatic, what can you say about the heat flow from b-c?

AM

Last edited by a moderator: May 5, 2017
3. Sep 19, 2011

osheari1

after working it out I ended up with tempurature values of T_a = 3.86K, T_b = 121.85, T_c = 30.88. however, these values seems extremely low, but I just assumed it was because of the very small volume.
I determined these values by calculated gamma. sense the gas is monatomic and ideal, 5/2R = C_p and C_p = C_v +R so Gamma = 5/2/(5/2-1) = 1.66

so, the heat flowing into the system is from the Isochoric process, Q = U + 0 = 3/2nRT = 1471.6J
the heat flowing out of the system is from the isobaric process, Q = 5/2nRT = -561.68J

The only work done by the process is from the adiabatic expansion and the isobaric process, so Wnet = Wbc + Wca = -(Ubc) + (Qca - Uca) = -(3/2*8.315(30.88-121.85)) + (-561.67 - 3/2nRT) = 1134.62 +(337 - 561.67) = 909.95 J

4. Sep 20, 2011

Andrew Mason

That looks right. You don't have to do so much calculation to find the work. By the first law, since there is no change in U in one cycle, so the work done by the gas must equal the net heat flow: $W = \Delta Q = Q_{in} - Q_{out} = 1471.6-561.7 = 909.9 J$

AM