If the ideal gas temperature at the steam point is `373.15K`, what is the limiting value of fhte ratio of pressure of a gas at lie steam point and at the triple point of water, when the gas is kept at constant volume?

Answer

To solve the problem, we need to determine the limiting value of the ratio of the pressure of an ideal gas at the steam point (373.15 K) to the pressure of the same gas at the triple point of water (273.16 K) when the gas is kept at constant volume.

### Step-by-Step Solution:

1. **Understand the Relationship**:
For an ideal gas at constant volume, the pressure (P) is directly proportional to the temperature (T). This relationship can be expressed as:
\[ P \propto T \]
This means that we can write:
\[ \frac{P_1}{P_2} = \frac{T_1}{T_2} \]
where \(P_1\) and \(P_2\) are the pressures at temperatures \(T_1\) and \(T_2\) respectively.

2. **Identify the Temperatures**:
- The temperature at the steam point (T1) is given as \(373.15 \, K\).
- The temperature at the triple point of water (T2) is given as \(273.16 \, K\).

3. **Set Up the Ratio**:
Using the relationship derived in step 1, we can set up the ratio of pressures:
\[ \frac{P_{\text{steam}}}{P_{\text{triple}}} = \frac{T_{\text{steam}}}{T_{\text{triple}}} \]
Substituting the known temperatures:
\[ \frac{P_{\text{steam}}}{P_{\text{triple}}} = \frac{373.15 \, K}{273.16 \, K} \]

4. **Calculate the Ratio**:
Now we compute the numerical value of the ratio:
\[ \frac{P_{\text{steam}}}{P_{\text{triple}}} = \frac{373.15}{273.16} \approx 1.366 \]

5. **Conclusion**:
The limiting value of the ratio of the pressure of the gas at the steam point to the pressure at the triple point of water is approximately \(1.366\).

### Final Answer:
\[ \text{Limiting value of } \frac{P_{\text{steam}}}{P_{\text{triple}}} \approx 1.366 \]

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