Efficiency of cyclic process

Homework Statement

(see attachment, ignore the arrows made with the pen)

The Attempt at a Solution

Efficiency of a cycle is defined as ##\eta=\frac{W}{Q}## where W is work done and Q is heat input.

W can be easily calculated by finding the area enclosed within the loop shown in the graph which is equal to ##(P-P_o)V_o##.

Heat input occurs only in the processes, B->A (##Q_1##) and B->C (##Q_2##).
##Q_1=nC_v\Delta T_1## where ##\Delta T_1=\frac{V_o}{nR}(P_o-P)##
##\Rightarrow Q_1=\frac{C_vV_o}{R}(P_o-P)##

##Q_2=nC_p\Delta T_2## and ##\Delta T_2=\frac{P_oV_o}{nR}##
##\Rightarrow Q_2=\frac{C_pP_oV_o}{R}##

$$\eta=\frac{W}{Q_1+Q_2}=\cfrac{(P_o-P)V_o}{\cfrac{C_vV_o}{R}(P_o-P)+\cfrac{C_pP_oV_o}{R}}$$
I don't have ##C_p## and ##C_v##, how am I supposed to solve this?

Any help is appreciated. Thanks!

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ehild
Homework Helper
What is the relation between Cp and Cv, and how does Cv depend on the degrees of freedom of the molecules in the gas? Assume monoatomic gas first.

ehild

What is the relation between Cp and Cv, and how does Cv depend on the degrees of freedom of the molecules in the gas? Assume monoatomic gas first.

ehild
The molar specific heat at constant volume is fR/2 and at constant pressure is (f/2+1)R where f is the degrees of freedom. For a monoatomic gas, f=3, hence Cv=3R/2 and Cp=5/2. Substituting the values give ##P=2P_o/7## which is not correct.

ehild
Homework Helper
f is not given, you get different pressures for different values of f. f=5 for diatomic gas and f=6 for three or more atomic molecules. For monoatomic gas, your solution should be correct.

ehild

f is not given, you get different pressures for different values of f. f=5 for diatomic gas and f=6 for three or more atomic molecules. For monoatomic gas, your solution should be correct.

ehild
I thought that it would be a bit difficult to put different values of f and find P. The answer in the key is A i.e ##0.3P_o##. I substituted it in the expression I ended up with and replaced ##C_v## and ##C_p## with fR/2 and (f/2+1)R respectively. Solving I got f=50/17. Is the question wrong?

ehild
Homework Helper
2/7 is 0.3 with one significant digit. As you had to choose among different data, you should have figured out which was possible.

ehild

2/7 is 0.3 with one significant digit. The question is wrong in the sense that it did not indicate the type of the gas. ((mono-atomic).

ehild
Thank you ehild!