InterviewSolution
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InterviewSolution
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Two ideal gas thermometer `A` and `B` use oxygen and hydrogen respectively . The following observations are made: Temperature,Pressure therometer A,Pressure therometer B Triple point of water,`1.250xx10^(5)Pa`,`0.200xx10^(5)Pa` Normal melting point of sulphur, ` 1.797xx10^(5)Pa`,`0.287xx10^(5)Pa` (a) What is the absolute temperature of normal melting point of sulphur as read by thermometers `A` and `B` ? (b) What do you think is the reason for the slightly different answers from `A` and `B` ? (The thermometers are not faulty). what further procedure is needed in the experiment to reduce the discrepancy between the two readings. |
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Answer» (a) Triple point of water, T = 273.16 K. At this temperature, pressure in thermometer `A, P_(A) = 1.250 xx 10^(5) `Pa Let `T_(1)` be the normal melting point of sulphur. At this temperature, pressure in thermometer` A, P_(1) = 1.797x× 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))=(1.797xx10^(5)xx273.16)/(1.250xx10^(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 ×x 10^(5)` Pa At temperature T1, the pressure in thermometer `B, P_(2) = 0.287 x× 10^(5)` Pa According to Charles’ law, we can write the relation: `(P_(B))/T=(P_(1))/(T_(1))` `(0.200xx10^(5))/(273.16)=(0.287xx10^(5))/(T_(1))` `:. T_(1)=(0.287xx10^(5))/(0.200xx10^(5))xx273.16=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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