A jar contains a gas and a few drops of water at `TK` The pressure in the jar is `830 mm` of Hg The temperature of the jar is reduced by `1%` The vapour pressure of water at two temperatures are 300 and 25 mm of Hg Calculate the new pressure in the jar .
(a)792mm of Hg
(b)817mm of Hg
(c)800mm of Hg
(d)840mm of Hg

Options

A. 792 mm of Hg
B. 817 mm of Hg
C. 800 mm of Hg
D. 840 mm of Hg

Explanation

To solve the problem step by step, we will follow the reasoning and calculations outlined in the video transcript.

### Step 1: Identify the initial conditions
- The initial total pressure in the jar (P1) is given as 830 mm of Hg.
- The vapor pressure of water at the initial temperature (TK) is not directly given, but we will assume it to be 30 mm of Hg based on the context provided in the transcript.

### Step 2: Calculate the pressure due to the gas at initial conditions
- The pressure due to the gas (Pg) can be calculated using the formula:
\[ Pg = P1 - P_{water} \]
Substituting the values:
\[ Pg = 830 \, \text{mm Hg} - 30 \, \text{mm Hg} = 800 \, \text{mm Hg} \]

### Step 3: Determine the new temperature after a 1% reduction
- The temperature is reduced by 1%. Therefore, the new temperature (T2) can be expressed as:
\[ T2 = T1 - 0.01 \cdot T1 = 0.99 \cdot T1 \]

### Step 4: Apply the gas law to find the new pressure
- Since the volume is constant, we can use the relationship between pressure and temperature:
\[ \frac{P1}{T1} = \frac{P2}{T2} \]
Rearranging gives:
\[ P2 = P1 \cdot \frac{T2}{T1} \]
Substituting for T2:
\[ P2 = P1 \cdot \frac{0.99 \cdot T1}{T1} = 0.99 \cdot P1 \]
Now substituting P1 (the pressure due to the gas):
\[ P2 = 0.99 \cdot 800 \, \text{mm Hg} = 792 \, \text{mm Hg} \]

### Step 5: Calculate the new saturated vapor pressure at the new temperature
- The vapor pressure of water at the new temperature (T2) is given as 25 mm of Hg.

### Step 6: Calculate the total pressure in the jar at the new temperature
- The total pressure in the jar (P_total) at the new conditions is the sum of the pressure due to the gas and the vapor pressure of water:
\[ P_{total} = P2 + P_{water} \]
Substituting the values:
\[ P_{total} = 792 \, \text{mm Hg} + 25 \, \text{mm Hg} = 817 \, \text{mm Hg} \]

### Final Answer
The new pressure in the jar is **817 mm of Hg**.