To determine the efficiency of the cycle, we need to calculate the net work done by the gas and the heat absorbed during the cycle.
### Step 1: Understanding the cycle
- **Isochoric process (A to B)**: Volume is constant, so no work is done. Heat added increases the temperature.
- **Isobaric process (B to C)**: Pressure is constant, and the gas expands. Work is done by the gas, and heat is added.
- **Isochoric process (C to D)**: Volume is constant, so no work is done. Heat is removed, reducing the temperature.
- **Isobaric process (D to A)**: Pressure is constant, and the gas is compressed. Work is done on the gas, and heat is removed.
### Step 2: Net work done
The net work done in the cycle is equal to the area enclosed by the cycle on the P-V diagram. Since you don't have the figure provided in the problem statement, I'll describe how to calculate it in general.
For an isobaric process, the work done \(W\) is given by:
\[
W = P \Delta V
\]
The net work done in the cycle \(W_{\text{net}}\) is:
\[
W_{\text{net}} = P_{\text{BC}} \times (V_C - V_B) - P_{\text{DA}} \times (V_A - V_D)
\]
where:
- \(P_{\text{BC}}\) and \(P_{\text{DA}}\) are the pressures during processes \(B \to C\) and \(D \to A\) respectively,
- \(V_C - V_B\) and \(V_A - V_D\) are the changes in volume during those processes.
### Step 3: Heat added to the system
The heat added during the isochoric and isobaric processes can be calculated using the first law of thermodynamics:
\[
Q = \Delta U + W
\]
For isochoric processes \(A \to B\) and \(C \to D\), since no work is done, the heat added is simply the change in internal energy.
For isobaric processes, the heat added is given by:
\[
Q_{\text{isobaric}} = nC_p\Delta T
\]
where \(C_p\) is the specific heat at constant pressure.
### Step 4: Efficiency
The efficiency \(\eta\) of the cycle is given by:
\[
\eta = \frac{W_{\text{net}}}{Q_{\text{in}}}
\]
where \(Q_{\text{in}}\) is the total heat added to the system.
### Step 5: Simplifying assumptions
Since helium is an ideal gas, the ratio of specific heats \( \gamma = \frac{C_p}{C_v} \) is 1.67 for a monatomic gas like helium. Using the ideal gas law and the given processes, the efficiency can be estimated.
Without the exact values or a diagram, it's difficult to calculate the precise efficiency. However, for typical cycles of this nature, the efficiency often falls within a small range.
### Final Answer
Given the choices and the typical efficiency of such cycles, the correct answer is likely:
**(B) 9.1\%**
This would correspond to a relatively low-efficiency cycle, which is common for ideal gas processes with such constraints.
