If at `60^(@)` and 80 cm of mercury pressure a definite masss of a gas is compressed slowly, then the final pressure of the gas if the final volume is half of the initial volume `(gamma =(3)/(2))` is

Options

A. a. 120 cm of Hg
B. b. 140 cm of Hg
C. c. 160 cm of Hg
D. d. 180 cm of Hg

Explanation

To solve the problem step by step, we will apply the principles of thermodynamics, specifically focusing on the isothermal process for an ideal gas.

### Step 1: Understand the Given Information
We have the following information:
- Initial temperature (T1) = 60°C (which remains constant)
- Initial pressure (P1) = 80 cm of mercury (Hg)
- Final volume (V2) is half of the initial volume (V1), so V2 = V1 / 2
- The process is isothermal (temperature remains constant)

### Step 2: Use the Ideal Gas Law for Isothermal Process
For an isothermal process, the relationship between pressure and volume for a given mass of gas can be expressed as:
\[ P_1 V_1 = P_2 V_2 \]

### Step 3: Substitute the Known Values
We know:
- \( P_1 = 80 \, \text{cm Hg} \)
- \( V_2 = \frac{V_1}{2} \)

Substituting these into the equation:
\[ 80 \, V_1 = P_2 \left( \frac{V_1}{2} \right) \]

### Step 4: Simplify the Equation
Now, we can simplify the equation:
\[ 80 \, V_1 = \frac{P_2 \, V_1}{2} \]

We can cancel \( V_1 \) from both sides (assuming \( V_1 \neq 0 \)):
\[ 80 = \frac{P_2}{2} \]

### Step 5: Solve for Final Pressure (P2)
Now, multiply both sides by 2 to solve for \( P_2 \):
\[ P_2 = 80 \times 2 \]
\[ P_2 = 160 \, \text{cm Hg} \]

### Final Answer
The final pressure of the gas after compression is:
\[ P_2 = 160 \, \text{cm Hg} \]