Two constant-volume gas thermometers are assembled, one with nitrogen and the other with hydrogen. Both contain enough gas so that.(a) What is the difference between the pressures in the two thermometers if both bulbs are in boiling water? (Hint: See Fig. 18-6) (b) Which gas is at higher pressure?

• 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:

$P_N=P_3\times \frac{T}{T_3}$

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

…….(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:
$P_H=P_3\times \frac{T}{T_3}$

Substitute the values and solve as:

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

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

$△P=P_N-P_H$

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

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: