A jar contains a gas and a few drops of water at absolute temperture `T_(1). The pressure in the jar is 830mmof mercuryand 25mmof mercury. The temperature of the jar is reduced by 1%. The saturation vapour pressures of water at the two temperatures are 30mm of murcury and 25mm of mercury. Calculate the new pressue in the jar.

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript while ensuring clarity and completeness in each step.

### Step 1: Identify Given Values
We are given:
- Initial total pressure in the jar, \( P_1 = 830 \, \text{mm of Hg} \)
- Initial saturation vapor pressure of water at temperature \( T_1 \), \( P_{sat1} = 30 \, \text{mm of Hg} \)
- New saturation vapor pressure of water after temperature reduction, \( P_{sat2} = 25 \, \text{mm of Hg} \)
- Temperature reduction of 1%, so the new temperature \( T_2 = 0.99 T_1 \)
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