Q. 54.1( 7 Votes )

# Two ideal gas the

Answer :

(a) We know that the triple point of water, T_{0} = 273.16 K.

At this temperature, pressure in thermometer A,

P_{A} = 1.250 × 10^{5} Pa

Let T be the normal melting point of sulphur.

At this temperature, pressure in thermometer A,

P = 1.797 × 10^{5} Pa

According to Charles’ law,

P_{A}/T_{0} = P/T

∴ T = (PT_{0})/P_{A}

⇒ T = (1.797×10^{5} Pa × 273.16 K)/(1.250×10^{5} Pa)

⇒ T = 392.69 K

Therefore, the absolute temperature corresponding to the normal melting point of sulphur according to the reading of thermometer A is 392.69 K.

At triple point T_{0} = 273.16 K, the pressure in thermometer B,

P_{B} = 0.200×10^{5} Pa

At temperature T, the pressure in thermometer B,

P’ = 0.287 × 10^{5} Pa

According to Charles’ law,

P_{B}/T_{0} = P’/T

⇒ (0.200 × 10^{5} Pa)/(273.16 K)= (0.287 × 10^{5} Pa)/T

∴ T = [(0.287 × 10^{5} Pa)/(0.200 × 10^{5} Pa)] × 273.16 K

⇒ T = 391.98 K

Therefore, the absolute temperature corresponding to the normal melting point of sulphur according to the reading of thermometer B is 391.98 K.

(b) The oxygen and hydrogen gases used in thermometers A and B respectively are not ideal gases. These gases have different physical behaviours. 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 ideal gases and will show same characteristics.

__NOTE:__ Ideal gases are those gases which follow ideal gas equation exactly.

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