Determine efficiency of carnot cycle if in adiabatic expansion volume 3 times of initial and r=1.5.

Options

A. `1-(1)/(sqrt(2))`
B. `1-(1)/(sqrt(3))`
C. `1+(1)/(sqrt(2))`
D. `1+(1)/(sqrt(3))`

Explanation

To determine the efficiency of a Carnot cycle given that the volume during adiabatic expansion is three times the initial volume and the ratio \( r = 1.5 \), we can follow these steps:

### Step 1: Understand the relationship in adiabatic processes
In an adiabatic process, the relationship between temperature and volume is given by:
\[ T V^{r-1} = \text{constant} \]
where \( r \) is the specific heat ratio.

### Step 2: Set up the equations for two states
For two states in the adiabatic process, we can write:
\[ T_1 V_1^{r-1} = T_2 V_2^{r-1} \]
where \( T_1 \) and \( V_1 \) are the initial temperature and volume, and \( T_2 \) and \( V_2 \) are the final temperature and volume.

### Step 3: Substitute the known values
Given that \( V_2 = 3 V_1 \) and \( r = 1.5 \):
\[ T_1 V_1^{1.5 - 1} = T_2 (3 V_1)^{1.5 - 1} \]
This simplifies to:
\[ T_1 V_1^{0.5} = T_2 (3 V_1)^{0.5} \]

### Step 4: Simplify the equation
We can simplify this further:
\[ T_1 = T_2 \cdot 3^{0.5} \]
or
\[ \frac{T_2}{T_1} = \frac{1}{\sqrt{3}} \]

### Step 5: Calculate the efficiency
The efficiency \( \eta \) of a Carnot cycle is given by:
\[ \eta = 1 - \frac{T_2}{T_1} \]
Substituting the value we found:
\[ \eta = 1 - \frac{1}{\sqrt{3}} \]

### Step 6: Final result
Thus, the efficiency of the Carnot cycle is:
\[ \eta = 1 - \frac{1}{\sqrt{3}} \]