NCERT solution Class 11 thermal properties of matter

NCERT Solutions

Class 11 Physics

Thermal Properties of Matter

NCERT Questions

Q. 10.1
Q. 10.2
Q. 10.3
Q. 10.4
Q. 10.5
Q. 10.6
Q. 10.7
Q. 10.8
Q. 10.9
Q. 10.10
Q. 10.11
Q. 10.12
Q. 10.13
Q. 10.14
Q. 10.15
Q. 10.16
Q. 10.17
Q. 10.18
Q. 10.19
Q. 10.20

Q. 10.5

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 × 10⁵ Pa 0.200 × 10⁵Pa

Normal melting point 1.797 × 10⁵Pa 0.287 × 10⁵ Pa

of sulphur

(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 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?

Given:

Triple point of water, T = 273.16 K.

At this temperature, pressure in thermometer A, P_A = 1.250 × 10⁵ Pa

Let T₁ be the normal melting point of sulphur.

At this temperature, pressure in thermometer A, P₁ = 1.797 x 10⁵ Pa

According to Charles’ law, we have the relation:

(P_A/T) = (P₁/T₁)

Therefore, T₁ = (P₁T)/ (P_A)

= (1.797 x 10⁵) x 273.16)/ (1.250 × 10⁵)

= 392.69 K

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

At triple point 273.16 K, the pressure in thermometer B, P_B = (0.200 × 10⁵) Pa

At temperature T₁, the pressure in thermometer B, P₂ = (0.287 × 10⁵) Pa

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

(P_B/T) = (P₁/T₁)

= (0.200 x 10⁵)/ (273.16) = (0.287 x 10⁵)/ (T₁)

Therefore T₁ = ((0.287 x 10⁵)/ (0.200 x 10⁵)) x (273.16)

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

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 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.

Video solution:

Thermal Properties of Matter Class 11 Physics NCERT Chapter 10 NCERT Solutions

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