The pressure of the gas in a constant volume gas thermometer is 80cm of mercury in melting ice at `1 atm`. When the bulb is placed in a liquid, the pressure becomes `160cm` of mercury. Find the temperature of the liquid.

Explanation

To solve the problem step by step, we will use the relationship between pressure and temperature in a constant volume gas thermometer.

### Step 1: Understand the relationship between pressure and temperature
In a constant volume gas thermometer, the pressure of the gas is directly proportional to its temperature. This can be expressed mathematically as:
\[ \frac{P_2}{P_1} = \frac{T_2}{T_1} \]
where:
- \( P_1 \) is the initial pressure (at melting ice),
- \( P_2 \) is the final pressure (in the liquid),
- \( T_1 \) is the initial temperature (at melting ice),
- \( T_2 \) is the final temperature (in the liquid).

### Step 2: Identify the known values
From the problem:
- \( P_1 = 80 \, \text{cm Hg} \)
- \( P_2 = 160 \, \text{cm Hg} \)
- \( T_1 = 273 \, \text{K} \) (temperature at melting ice)

### Step 3: Substitute the known values into the equation
We can rearrange the equation to solve for \( T_2 \):
\[ T_2 = T_1 \cdot \frac{P_2}{P_1} \]

### Step 4: Plug in the values
Now substituting the known values into the equation:
\[ T_2 = 273 \, \text{K} \cdot \frac{160 \, \text{cm Hg}}{80 \, \text{cm Hg}} \]

### Step 5: Simplify the equation
Calculating the fraction:
\[ \frac{160}{80} = 2 \]
So, the equation becomes:
\[ T_2 = 273 \, \text{K} \cdot 2 \]

### Step 6: Calculate \( T_2 \)
Now, calculate \( T_2 \):
\[ T_2 = 546 \, \text{K} \]

### Final Answer
The temperature of the liquid is \( 546 \, \text{K} \).

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