Two glass bulbs of same volume containing a gas at NTP are connected by a very narrow tube. The pressure is 88.46cm of mercury when one bulb is kept in ice and the other in hot water. Calculate the temperature of the hot water.

Explanation

To solve the problem step by step, we will use the ideal gas law and the relationship between pressure, volume, and temperature.

### Step-by-Step Solution:

1. **Understand the Given Conditions:**
- We have two glass bulbs of the same volume containing a gas at Normal Temperature and Pressure (NTP).
- The pressure in the system is given as 88.46 cm of mercury (Hg) when one bulb is in ice and the other in hot water.
- At NTP, the pressure (P1) is 76 cm of Hg and the temperature (T1) is 20°C, which is 293 K when converted to Kelvin.

2. **Set Up the Relation Using the Ideal Gas Law:**
- Since the volume of the bulbs remains constant, we can use the relation:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
- Here, \(P_1\) is the pressure at NTP, \(T_1\) is the temperature at NTP, \(P_2\) is the pressure when one bulb is in ice and the other in hot water, and \(T_2\) is the temperature of the hot water (which we need to find).

3. **Substitute the Known Values:**
- From the problem, we have:
- \(P_1 = 76 \, \text{cm Hg}\)
- \(T_1 = 293 \, \text{K}\)
- \(P_2 = 88.46 \, \text{cm Hg}\)
- We need to find \(T_2\) (the temperature of the hot water).

4. **Rearranging the Equation:**
- Rearranging the equation gives us:
\[ T_2 = \frac{P_2 \cdot T_1}{P_1} \]

5. **Plugging in the Values:**
- Substitute the values into the equation:
\[ T_2 = \frac{88.46 \, \text{cm Hg} \cdot 293 \, \text{K}}{76 \, \text{cm Hg}} \]

6. **Calculate \(T_2\):**
- Performing the calculation:
\[ T_2 = \frac{88.46 \times 293}{76} \approx 341.03 \, \text{K} \]

7. **Convert Kelvin to Celsius:**
- To convert the temperature from Kelvin to Celsius:
\[ T_{Celsius} = T_2 - 273 = 341.03 - 273 \approx 68.03 \, °C \]

### Final Answer:
The temperature of the hot water is approximately **68.03 °C**.