The pressure of the gas in a constant volume gas thermometer is 70 kPa at the ice point. Find the pressure at the steam point.

#### Solution

Given:

Temperature of ice point, T_{1} = 273.15 K

Temperature of steam point, T_{2} = 373.15 K

Pressure of the gas in a constant volume thermometer at the ice point, P_{1}_{} = 70 kPa,

Let P_{tr}_{ }be the pressure at the triple point and P_{2}_{ }be the pressure at the steam point.

The temperature-pressure relations for ice point and steam point are given below:

For ice point,

\[T_1 = \frac{P_1}{P_{tr}} \times 273 . 16 K\]

\[\Rightarrow 273 . 15 = \frac{70}{P_{tr}} \times {10}^3 \times 273 . 16\]

\[ \Rightarrow P_{tr} = \frac{70 \times 273 . 16 \times {10}^3}{273 . 15} Pa\]

For steam point,

\[T_2 = \frac{P_2 \times 273 . 16}{P_{tr}} K\]

On substituting the value of P_{tr} ,we get:

\[373 . 15 = \frac{P_2 \times 273 . 15 \times 273 . 16}{70 \times 273 . 16 \times {10}^3} \]

\[ \Rightarrow P_2 = \frac{373 . 15 \times 70 \times {10}^3}{273 . 15}\]

\[ \Rightarrow P_2 = 95 . 626 \times {10}^3 Pa\]

\[ \Rightarrow P_2 \simeq 96 kPa\]

Therefore, the pressure at steam point is 96 kPa.