__Question 11.5__:Temperature |
Pressure
thermometer A |
Pressure
thermometer B |

Triple-point of water | 1.250 ×
10^{5}
Pa |
0.200 ×
10^{5}
Pa |

Normal melting point of sulphur | 1.797 ×
10^{5}
Pa |
0.287 ×
10^{5}
Pa |

*?*

(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?

__Solution__:**(a)**Triple point of water,

*T*= 273.16 K.

At this temperature, pressure in thermometer A,

*P*

_{A}= 1.250 × 10

^{5}Pa

Let

*T*

_{1}be the normal melting point of sulphur.

At this temperature, pressure in thermometer A,

*P*

_{1}= 1.797 × 10

^{5}Pa

According to Charles’ law, we have the relation:

P

_{A}/ T = P

_{1}/ T

_{1}

∴ T

_{1}= (P

_{1}T) / P

_{}

_{A}= (1.797 X 10 X 273.16) / (1.250 X 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.

At triple point 273.16 K, the pressure in thermometer B,

*P*

_{B}= 0.200 × 10

^{5}Pa

At temperature

*T*

_{1}, the pressure in thermometer B,

*P*

_{2}= 0.287 × 10

^{5}Pa

According to Charles’ law, we can write the relation:

P

_{B}/ T = P

_{1}/ T

_{1}

(0.200 X 10

^{5}) / 273.16 = (0.287 X 10

^{5}) / T

^{}

_{1}

∴ T

_{1}= [ (0.287 X 10

^{5}) / (0.200 X 10

^{5}) ] X 273.16 = 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 oxygen and hydrogen gas present in thermometers A and B respectively are not perfect ideal gases. Hence, there is a slight difference between the readings of thermometers A and B.

To reduce the discrepancy between the two readings, the experiment should be carried under low pressure conditions. At low pressure, these gases behave as perfect ideal gases.

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