Question

The pressure measured by a constant volume gas thermometer is 40 kPa at the triple point of water. What will be the pressure measured at the boiling point of water (100°C)?

Answer 1

The pressure measured at the boiling point of water (100°C) is $\mathrm{p} \approx 55 \mathrm{kPa}$.

Explanation:  

Step 1:

Given,

Let the temperature of water at boiling point is represented by “T”

T(temperature at boiling point of water)= $100^{\circ} \mathrm{C}=273+100=373 \mathrm{K}$

Pressure at triple point of water = 40 kPa

Let the triple point pressure of water is represented as “Pt”

Step 2:

By using the formula  

$\mathrm{T}=\frac{p}{p_{t}} \times 273.16$

$\mathrm{P}=\frac{T}{273.16} \times P_{t}$

$P=\frac{373}{273.16} \times 40 \times 10^{3}$

Step 3:

By solving the above equation                                  

$\mathrm{P}=54.62 \times 10^{3}$

Approximate value is

$\mathrm{p} \approx 55 \mathrm{kPa}$

Answer 2

Answer:

the boiling point of water (100°C) is \mathrm{p} \approx 55 \mathrm{kPa}p≈55kPa .

Explanation:  

Step 1:

Given,

Let the temperature of water at boiling point is represented by “T”

T(temperature at boiling point of water)= 100^{\circ} \mathrm{C}=273+100=373 \mathrm{K}100∘C=273+100=373K

Pressure at triple point of water = 40 kPa

Let the triple point pressure of water is represented as “Pt”

Step 2:

By using the formula  

\mathrm{T}=\frac{p}{p_{t}} \times 273.16T=ptp×273.16

\mathrm{P}=\frac{T}{273.16} \times P_{t}P=273.16T×Pt

P=\frac{373}{273.16} \times 40 \times 10^{3}P=273.16373×40×103

Step 3:

By solving the above equation                                  

\mathrm{P}=54.62 \times 10^{3}P=54.62×103

Approximate value is

\mathrm{p} \approx 55 \mathrm{kPa}p≈55kPa