Question

Two ideal gas thermometers Aand Buse oxygen and hydrogen respectively. The following observations are made:




Temperature Pressure thermometer A Pressure thermometer B


Triple-point of water 1.250 × 105 Pa 0.200 × 105 Pa

Normal melting point of sulphur 1.797 × 105 Pa 0.287 × 105 Pa




(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers Aand B?

(b) What do you think is the reason behind the slight difference in answers of thermometers Aand B? (The thermometers are not faulty). What further procedure is needed in

the experiment to reduce the discrepancy between the two readings?

Answer 1

Let the melting point of Sulphur is T( in K)
for thermometer A,
pressure at triple point of water (Po) = 1.25 × 10^5 Pa
Triple point of water( To) = 273.16K
Pressure at normal point of Sulphur (P) = 1.797 × 10^5 pa
Now,
use relation,
P1/T1 = P2/T2 [ V = constant from gaseous Law ]
Po/To = P/T
1.250 × 10^5/273.16 = 1797× 10^5/T
T = 1.797/1.250 × 273.16
= 392.69 K

Similarly for thermometer B,
Po = 0.200 × 10^5 Pa
P = 0.287 × 10^5 Pa
To = 273.16 K
T = ?
T = Po/P × To
= 0.200/0.287 × 273.16
= 391.98K


(b) actually Hydrogen gas and Oxygen are not perfectly ideal gas , this formula [ P1/T1 = P2/T2] is taken as ideal gas. So, slightly different answers of thermometer A and B .