Question

The pressure of the gas in a constant volume gas thermometer is 70kpa at the ice point. find the pressure at the steam point ​

Answer 1

Explanation:

Temperature of ice point T1 = 273.15 K

Temperature of steam point T2 = 372.15 K

Pressure of the gas in a constant volume thermometer at ice point P1 = 70 kpa

Now,

Let Ptr be the pressure of the triple point and P2 be the pressure at the steam point.

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

For Ice Point,

T1 = P1/Ptr × 273.16 K

273.15 = 70/Ptr × 10×10×10×273.16

Ptr = (70×10×10×10×273.16)/273.15 Pa

For Steam Point,

T2 = 273.16/Ptr K

On substituting the value of Ptr, we get:

373.15 = (P2 × 273.15 × 273.16)/(70 × 273.16 × 10 × 10 × 10)

P2 = 95.626 × 1000 Pa

P2 = 96kpa (approx.)

Therefore, the pressure at steam point is 96kpa

Answer 2

The Pressure at the stream point is $\mathrm{p} \approx 96 \mathrm{kPa}$

Explanation:

Step 1:

By using the formula

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

Step 2:  

(a) For ice point $T=273 \mathrm{K}$, $\mathrm{P}=70 \mathrm{kPa}$

$273=\frac{70 \times 10^{3}}{P_{t}} \times 273.16$

$P_{t}=70 \times 10^{3} \times \frac{273.16}{273}$   ………………………………….(1)

Step 3:

(b) For steam point $T=373 \mathrm{K}$

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

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

Step 4:

By using the equation (1)

$\mathrm{P}=373 \times 70 \times 10^{3} \times \frac{273.16}{273.16 \times 273}$

By doing calculation in above equation we get                                            

$\mathrm{p}=95.64 \times 10^{3}$

Approximate value of pressure is

$\mathrm{p} \approx 96 \mathrm{kPa}$