91Ó°ÊÓ

• i) For both gases, pressure is $P_3=80kPa$
• ii) Two constant-volume of gases, one with nitrogen and the other with hydrogen

By using the concept of a constant volume gas thermometer, find the difference between the pressures in the two thermometers when kept in boiling water and which gas is at greater pressure.

From Equation 18-6, the temperature T as measured with a gas thermometer is given by:
$T=(273.16K)\left(\frac{P_N}{P_3}\right)..............\left(1\right)$

For nitrogen thermometer, the temperature is given as:
$T_N=373.35K$

$Considerforpressureof{P}_3=80k{P}_a$
$$\begin{gathered} T_3=Triplepointtemperature \\ =273.16 \end{gathered}$$

Now using all these values for nitrogen in equation 1, solve for the pressure by the nitrogen gas as:

$P_N=P_3×\frac{T}{T_3}$

Substitute the values and solve as:
localid="1661341151894" $$\begin{gathered} P_N=80×\frac{373.35}{273.16 \\ } \\ =109.343k{P}_a \end{gathered}$$

…â¶Ä¦.(2)

Similarly, for hydrogen thermometer consider the parameters as:

$$\begin{gathered} T_H=373.15K \\ {P}_3=80k{P}_a \\ {T}_3=273.16 \end{gathered}$$

Using in relation of equation (1), solve for pressure by the hydrogen gas as:
$P_H=P_3×\frac{T}{T_3}$

Substitute the values and solve as:

$$\begin{gathered} P_H=80×\frac{375.15}{273.16} \\ =109.284k{P}_a \end{gathered}$$ ….. (3)

Subtracting equation 2 and 3, we get the pressure difference as given:

$â–³P=P_N-P_H$

Substitute the values and solve as:
$$\begin{gathered} △P=109.343-109.284 \\ =0.059kPa \\ ≈0.06kPa \end{gathered}$$

Hence, the pressure difference between nitrogen and hydrogen is 0.06 kPa.

From equation (2) and (3) in the part (a) calculations it is observed as follows: