Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made: 

Temperature Pressure thermometer A Pressure thermometer B
Triple-point of water  1.250 × 105Pa 0.200 × 105Pa
Normal melting point of sulphur 1.797 × 105Pa 0.287 × 105 Pa

 

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

(b) What do you think is the reason behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

Asked by Pragya Singh | 1 year ago |  78

1 Answer

Solution :-

(a) Triple point of water, T= 273.16K.

Pressure in thermometer A at the  triple point, PA =1.25×105 Pa

Normal melting point of sulphur = T1

Pressure in thermometer A at this temperature, P1=1.797×105 Pa

According to Charles law, we have the relation:

\( \dfrac{P_A}{T}=\dfrac{P_1}{T_1}\)

\(T_1= \dfrac{P_1 T}{P_A}\)

\( \dfrac{(1.797×10^5×273.16)}{1.25×10^5}\)

=392.69 K

Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer A is 392.69 K.

 

(b) The pressure in thermometer B at the triple point of water, PB =0.2×10Pa

The temperature in thermometer B at the normal melting point (T1) of sulphur is, 

P1 =  0.287×105 Pa

According to Charles law, we can write the relation:
\( \dfrac{P_B}{T}=\dfrac{P_1}{T_1}\)

​​\( T_1= \dfrac{P_1 T}{P_B}\)

T\( \dfrac{ (0.287×10^5 × 273.16)}{0.2×10^5}\)
=391.98 K

Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer B is 391.98 K.

(b) The reason for the discrepancy is that the gases do not behave like ideal gas in practice. The discrepancy can be reduced by taking the reading at very low pressure so that the gases show perfect behaviour.

Answered by Abhisek | 1 year ago

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