The fall in temperature of helium gas initially at `20^(@)` when it is suddenly expanded to 8 times its original volume is `(gamma=(5)/(3))`

Options

A. a.70.25 K
B. b.71.25 K
C. c.72.25 K
D. d.73.25 K

Explanation

To solve the problem of the fall in temperature of helium gas when it is suddenly expanded to 8 times its original volume, we will use the principles of thermodynamics, specifically the adiabatic process. Here’s a step-by-step solution:

### Step 1: Identify the initial conditions
The initial temperature \( T_1 \) of the helium gas is given as \( 20^\circ C \). To work in Kelvin, we convert this:
\[ T_1 = 20 + 273 = 293 \, K \]

### Step 2: Understand the adiabatic process
In an adiabatic process, the relationship between temperature and volume is given by:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
where \( \gamma \) (gamma) is the heat capacity ratio, given as \( \frac{5}{3} \).

### Step 3: Define the volumes
Let \( V_1 \) be the initial volume. The final volume \( V_2 \) after expansion is:
\[ V_2 = 8 V_1 \]

### Step 4: Substitute values into the adiabatic equation
Now, substituting \( V_2 \) into the adiabatic equation:
\[ T_1 V_1^{\gamma - 1} = T_2 (8 V_1)^{\gamma - 1} \]

### Step 5: Simplify the equation
We can simplify the equation:
\[ T_1 V_1^{\gamma - 1} = T_2 \cdot 8^{\gamma - 1} V_1^{\gamma - 1} \]
Dividing both sides by \( V_1^{\gamma - 1} \):
\[ T_1 = T_2 \cdot 8^{\gamma - 1} \]

### Step 6: Solve for \( T_2 \)
Rearranging the equation to solve for \( T_2 \):
\[ T_2 = \frac{T_1}{8^{\gamma - 1}} \]

### Step 7: Calculate \( \gamma - 1 \)
First, calculate \( \gamma - 1 \):
\[ \gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3} \]

### Step 8: Calculate \( 8^{\gamma - 1} \)
Now calculate \( 8^{\frac{2}{3}} \):
\[ 8^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^2 = 4 \]

### Step 9: Substitute back to find \( T_2 \)
Now substitute back into the equation for \( T_2 \):
\[ T_2 = \frac{293}{4} = 73.25 \, K \]

### Step 10: Conclusion
The final temperature \( T_2 \) after the sudden expansion is:
\[ T_2 = 73.25 \, K \]