NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter

Ans : \(\begin{array}{l}{\text { Kelvin and Celsius scales are related as: }} \\ {T_{\mathrm{c}}=T_{\mathrm{K}}-273.15 \ldots(i)} \\ {\text { Celsius and Fahrenheit scales are related as: }} \\ {T_{\mathrm{F}}=\frac{9}{5} T_{\mathrm{C}}+32}\end{array}\)
\(\begin{array}{l}{\text { Celsius and Fahrenheit scales are related as: }} \\ {T_{\mathrm{F}}=\frac{9}{5} T_{\mathrm{C}}+32}\end{array}\)
\(\begin{array}{l}{\frac{\text { For neon: }}{T_{k}=24.57 \mathrm{K}}} \\ {\therefore T_{\mathrm{c}}=24.57-273.15=-248.58^{\circ} \mathrm{C}} \\ {T_{\mathrm{F}}=\frac{9}{5} T_{\mathrm{C}}+32}\end{array}\)
\(\begin{array}{l}{=\frac{9}{5}(-248.58)+32} \\ {=415.44^{\circ} \mathrm{F}}\end{array}\)
\(\begin{array}{l}{\text { For carbon dioxide: }} \\ {T_{k}=216.55 \mathrm{K}} \\ {\therefore T_{\mathrm{c}}=216.55-273.15=-56.60^{\circ} \mathrm{C}} \\ {T_{\mathrm{F}}=\frac{9}{5}\left(T_{\mathrm{C}}\right)+32}\end{array}\)
\(\begin{array}{l}{=\frac{9}{5}(-56.60)+32} \\ {=-69.88^{\circ} \mathrm{C}}\end{array}\) 

Ans : \(\begin{array}{l}{\text { Triple point of water on absolute scaleA, } T_{1}=200 \mathrm{A}} \\ {\text { Triple point of water on absolute scale } \mathrm{B}, T_{2}=350 \mathrm{B}} \\ {\text { Triple point of water on Kelvin scale, } T_{\kappa}=273.15 \mathrm{K}} \\ {\text { The temperature } 273.15 \mathrm{K} \text { on Kelvin scale is equivalent to } 200 \mathrm{A} \text { on absolute scale } \mathrm{A} \text { . }}\end{array}\)
\(\begin{array}{l}{T_{1}=T_{\mathrm{k}}} \\ {200 \mathrm{A}=273.15 \mathrm{K}} \\ {\therefore \mathrm{A}=\frac{273.15}{200}} \\ {\text { The temperature } 273.15 \mathrm{K} \text { on Kelvin scale is equivalent to } 350 \mathrm{B} \text { on absolute scale B. }}\end{array}\)
\(\begin{array}{l}{T_{2}=T_{\kappa}} \\ {350 \mathrm{B}=273.15} \\ {\therefore \mathrm{B}=\frac{273.15}{350}} \\ {T_{\mathrm{A}} \text { is triple point of water on scale } \mathrm{A} \text { . }} \\ {T_{\mathrm{B}} \text { is triple point of water on scale B. }}\end{array}\)
\(\begin{array}{l}{\therefore \frac{273.15}{200} \times T_{\mathrm{A}}=\frac{273.15}{350} \times T_{\mathrm{B}}} \\ {T_{\mathrm{A}}=\frac{200}{350} T_{\mathrm{B}}} \\ {\text { Therefore, the ratio } T_{\mathrm{A}} : T_{\mathrm{B}} \text { is given as } 4 : 7 .}\end{array}\) 

Ans : \(\begin{array}{l}{\text { It is given that: }} \\ {R=R_{0}\left[1+\mathrm{a}\left(T-T_{0}\right)\right] \ldots(i)} \\ {\text { Where, }} \\ {R_{0} \text { and } T_{0} \text { are the initial resistance and temperature respectively }} \\ {R \text { and } T \text { are the final resistance and temperature respectively }} \\ {\text { a is a constant }}\end{array}\)
\(\begin{array}{l}{\text { At the triple point of water, } T_{0}=273.15 \mathrm{K}} \\ {\text { Resistance of lead, } R_{0}=101.6 \Omega} \\ {\text { At normal melting point of lead, } T=600.5 \mathrm{k}} \\ {\text { Resistance of lead, } R=165.5 \Omega} \\ {\text { Substituting these values in equation (i), we get: }}\end{array}\)
\(\begin{array}{l}{R=R_{0}\left[1+\alpha\left(T-T_{0}\right)\right]} \\ {165.5=101.6[1+\alpha(600.5-273.15)]} \\ {1.629=1+\alpha(327.35)} \\ {\therefore \alpha=\frac{0.629}{327.35}=1.92 \times 10^{-3} \mathrm{K}^{-1}} \\ {\text { For resistance, } R_{1}=123.4 \Omega} \\ {\mathrm{R}_{1}=\mathrm{R}_{0}\left[1+\alpha\left(\mathrm{T}-\mathrm{T}_{0}\right)\right]}\end{array}\)
\(\begin{array}{l}{\text { Where, T is the temperature when the resis tan ce of lead is } 123.4 \Omega} \\ {123.4=101.6\left[1+1.92 \times 10^{-3}(\mathrm{T}-273.15)\right]} \\ {1.214=1+1.92 \times 10^{-3}(\mathrm{T}-273.15)} \\ {\frac{0.214}{1.92 \times 10^{-3}}=\mathrm{T}-273.15} \\ {\therefore \mathrm{T}=384,61 \mathrm{K}}\end{array}\) 

Ans : \(\begin{array}{l}{\text { (a) The triple point of water has a unique value of } 273.16 \mathrm{K} \text { . At particular values of }} \\ {\text { volume and pressure, the triple point of water is always } 273.16 \mathrm{K} \text { . The melting point of }} \\ {\text { ice and bolling point of water do not have particular values because these points depend }} \\ {\text { on pressure and temperature. }}\end{array}\)
(b) The absolute zero or O K is the other fixed point on the Kelvin absolute scale.
\(\begin{array}{l}{\text { (c) The temperature } 273.16 \mathrm{K} \text { is the triple point of water. It is not the melting point of }} \\ {\text { ice. The temperature } 0^{\circ} \mathrm{C} \text { on Celsius scale is the melting point of ice. Its corresponding }} \\ {\text { value on Kelvin scale is } 273.15 \mathrm{K} \text { . }} \\ {\text { Hence, absolute temperature (Kelvin scale) } T, \text { is related to temperature } t_{c \prime} \text { on Celsius }} \\ {t_{c}=T-273.15}\end{array}\)
\(\begin{array}{l}{\text { (d) Let } T_{F} \text { be the temperature on Fahrenheit scale and } T_{\kappa} \text { be the temperature on }} \\ {\text { absolute scale. Both the temperatures can be related as: }} \\ {\frac{T_{F}-32}{180}=\frac{T_{K}-273.15}{100}}\end{array}\)
\(\begin{array}{l}{\text { Let } T_{F_{1}} \text { be the temperature on Fahrenheit scale and } T_{\kappa 1} \text { be the temperature on absolute }} \\ {\text { scale. Both the temperatures can be related as: }} \\ {\frac{T_{r 1}-32}{180}=\frac{T_{\mathrm{k} 1}-273.15}{100}}\end{array}\)
\(\begin{array}{l}{\text { It is given that: }} \\ {T_{\mathrm{k} 1}-T_{\mathrm{k}}=1 \mathrm{K}}\end{array}\)
\(\begin{array}{l}{\text { Subtracting equation (i) from equation (ii), we get: }} \\ {\frac{T_{\mathrm{FI}}-T_{\mathrm{F}}}{180}=\frac{T_{\mathrm{KI}}-T_{\mathrm{K}}}{100}=\frac{1}{100}}\end{array}\)
\(\begin{array}{l}{T_{\mathrm{F} 1}-T_{\mathrm{F}}=\frac{1 \times 180}{100}=\frac{9}{5}} \\ {\text { Triple point of water }=273.16 \mathrm{K}} \\ {\text { ¿Triple point of water on absolute scale }=273.16 \times \frac{9}{5}=491.69}\end{array}\) 

Ans : \(\begin{array}{l}{\text { (a) Triple point of water, } T=273.16 \mathrm{K} \text { . }} \\ {\text { At this temperature, pressure in thermometer } A, P_{A}=1.250 \times 10^{5} \text { Pa }} \\ {\text { Let } T_{1} \text { be the normal melting point of sulphur. }} \\ {\text { At this temperature, pressure in thermometer } A_{r} P_{1}=1.797 \times 10^{5} \text { Pa }} \\ {\text { According to Charles' law, we have the relation: }}\end{array}\)
\(\begin{array}{l}{\frac{P_{\mathrm{A}}}{T}=\frac{P_{1}}{T_{1}}} \\ {\therefore T_{1}=\frac{P_{1} T}{P_{\mathrm{A}}}=\frac{1.797 \times 10^{5} \times 273.16}{1.250 \times 10^{5}}} \\ {=392.69 \mathrm{K}} \\ {\text { Therefore, the absolute temperature of the normal melting point of sulphur as read by }} \\ {\text { thermometer } \mathrm{A} \text { is } 392.69 \mathrm{K} \text { . }}\end{array}\)
\(\begin{array}{l}{\text { Therefore, the absolute temperature of the normal melting point of sulphur as read by }} \\ {\text { thermometer } A \text { is } 392.69 \mathrm{K} \text { . }} \\ {\text { At triple point } 273.16 \mathrm{K} \text { , the pressure in thermometer } \mathrm{B}, P_{\mathrm{B}}=0.200 \times 10^{5} \mathrm{Pa}} \\ {\text { At temperature } T_{1}, \text { the pressure in thermometer } \mathrm{B}, P_{2}=0.287 \times 10^{5} \mathrm{Pa}} \\ {\text { According to Charles' law, we can write the relation: }}\end{array}\)
\(\begin{array}{l}{\frac{P_{\mathrm{B}}}{T}=\frac{P_{1}}{T_{1}}} \\ {\frac{0.200 \times 10^{5}}{273.16}=\frac{0.287 \times 10^{3}}{T_{1}}} \\ {\therefore T_{1}=\frac{0.287 \times 10^{5}}{0.200 \times 10^{5}} \times 273.16=391.98 \mathrm{K}}\end{array}\)
\(\begin{array}{l}{\text { Therefore, the absolute temperature of the normal melting point of sulphur as read by }} \\ {\text { thermometer } B \text { is } 391.98 \mathrm{K} \text { . }}\end{array}\)
\(\begin{array}{l}{\text { (b) The oxygen and hydrogen gas present in thermometers } A \text { and } B \text { respectively are not }} \\ {\text { perfect ideal gases. Hence, there is a slight difference between the readings of }} \\ {\text { thermometers } A \text { and } B \text { . }}\end{array}\)
\(\begin{array}{l}{\text { To reduce the discrepancy between the two readings, the experiment should be carried }} \\ {\text { under low pressure conditions. At low pressure, these gases behave as perfect ideal }} \\ {\text { gases. }}\end{array}\)