Thermal Properties of Matter

Class 11 NCERT Physics

NCERT

1   The triple points of neon and carbon dioxide are $24.57 K$ and $216.55 K$ respectively. Express these temperatures on the Celsius and Fahrenheit scales.

Solution :

Kelvin and Celsius scales are related as:$\\$ $T_C=T_K-273.15....(i)$$\\$ Celsius and Fahrenheit scales are related as:$\\$ $T_F=\dfrac{9}{5}T_C+32....(ii)$$\\$ $\underline{For \ neon:}\\ T_K=24.57 K\\ \therefore T_C=24.57-273.15=-248.58^oC\\ T_F=\dfrac{9}{5}T_C+32\\ =\dfrac{9}{5}(-248.58)+32\\ =415.44^o F\\ \underline{For \ carbon \ dioxide:}\\ T_k=216.55 K\\ \therefore T_C=216.55-273.15=-56.60^oC.\\ T_F=\dfrac{9}{5}T_C+32\\ =\dfrac{9}{5}(-56.60)+32\\ =-69.88^o C$

2   Two absolute scales A and B have triple points of water defined to be $200$ A and $350$ B. What is the relation between $T_A$ and $T_B $?

Solution :

Triple point of water on absolute scale A, $T_1 =200$ A .$\\$ Triple point of water on absolute scale B, $T_2 = 350$ B Triple$\\$ point of water on Kelvin scale, $T_K = 273.15 K$$\\$ The temperature $273.15 K$ on Kelvin scale is equivalent to $200$ A on absolutes scale A.$\\$ $T_1 = T_K\\ 200 A = 273.15 K\\ \therefore A=\dfrac{273.15}{200}$$\\$ The temperature $273.15 K$ on Kelvin scale is equivalent to $350$ B on absolute scale B.$\\$ $T_2 = T_K\\ 350 B=273.15\\ \therefore B=\dfrac{273.15}{350}$$\\$ $T_A$ is triple point of water on scale $A.$$\\$ $T_ B$ is triple point of water on scale $B.$$\\$ $\therefore \dfrac{273.15}{200}*T_A\\ =\dfrac{273.15}{350}*T_B\\ T_A=\dfrac{200}{350}T_B$$\\$ Therefore, the ratio $T_A : T_B$ is given as $4 : 7.$

3   The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law $R = R_o[ 1 +\alpha (T - T_o )]$ The resistance is $101.6 \Omega$ at the triple-point of water $273.16 K,$ and $165.5 \Omega$ at the normal melting point of lead $(600.5 K)$. What is the temperature when the resistance is $123.4 \Omega$?

Solution :

It is given that:$\\$ $R = R_o[ 1 +\alpha (T - T_o )].......(i)$$\\$ Where,$\\$ $R_0$ and $T_0$ are the initial resistance and temperature respectively $R$ and $T$ are the final resistance and temperature respectively $\alpha $ is a constant$\\$ At the triple point of water, $T_0 = 273.15 K$$\\$ Resistance of lead, $R_0 = 101.6 \Omega $$\\$ At normal melting point of lead, $T = 600.5 K$$\\$ Resistance of lead, $R = 165.5 \Omega$$\\$ Substituting these values in equation (i), we get:$\\$ $R = R_o[ 1 +\alpha (T - T_o )]\\ 165.5 = 101.6 [ 1 + a ( 600.5 - 273.15 )]\\ 1.629 = 1 +\alpha( 327.35)\\ \therefore \alpha =\dfrac{0.629}{327.35}\\ =1.92*10^{-3}K(-1)$$\\$ For resistance,$R_1=123.4\Omega\\ R_1=R_0[1+\alpha(T-T_0)]$$\\$ Where, $T$ is the temperature when the resistance of lead is $123.4\Omega $$\\$ $123.4 = 101.6 [ 1 + 1.92 * 10^{-3} ( T - 273.15)]\\ 1.214 = 1 + 1.92 * 10^{-3} (T - 273.15)\\ \dfrac{0.214}{1.92*10^{-3}} \\ =T-273.15\\ \therefore T=384.61K$

4   Answer the following:$\\$ (a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?$\\$ (b)There were two fixed points in the original Celsius scale as mentioned above which were assigned the number $0^oC$ and $100^oC$ respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number $273.16 K.$ What is the other fixed point on this (Kelvin) scale? The absolute temperature (Kelvin scale) $T$ is related to the temperature $t_c$ on the Celsius scale by $\\$ $t_c = T - 273.15$$\\$ Why do we have $273.15$ in this relation, and not $273.16$?$\\$ (d)What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

Solution :

(a)The triple point of water has a unique value of $273.16 K.$ At particular values of volume and pressure, the triple point of water is always $273.16 K.$ The melting point of ice and boiling point of water do not have particular values because these points depend on pressure and temperature.$\\$ (b)The absolute zero or $0 K$ is the other fixed point on the Kelvin absolute scale.$\\$ (c)The temperature $273.16 K$ is the triple point of water. It is not the melting point of ice. The temperature $0^oC$ on Celsius scale is the melting point of ice. Its corresponding value on Kelvin scale is $273.15 K.$ Hence, absolute temperature (Kelvin scale) $T$ , is related to temperature $t_c$ , on Celsius scale as:$\\$ $t_c = T - 273.15$$\\$ (d)Let $T_F$ be the temperature on Fahrenheit scale and $T_K$ be the temperature on absolute scale. Both the temperatures can be related as:$\\$ $\dfrac{T_F-32}{180}=\dfrac{T_K-273.15}{100}....(i)$$\\$ Let $T_{ F1}$ be the temperature on Fahrenheit scale and $T_{K1}$ be the temperature on absolute scale. Both the temperatures can be related as:$\\$ $\dfrac{T_{F1}-32}{180}=\dfrac{T_{KI}-273.15}{100}...(ii)$$\\$ It is given that:$\\$ $T_{K1}-T_K=1 K$$\\$ Subtracting equation (i) from equation (ii), we get:$\\$ $\dfrac{T_{F1}-T_F}{180}=\dfrac{T_{K1}-T_K}{100}=\dfrac{9}{5}$$\\$ Triple point of water $= 273.16 K$$\\$ $\therefore $ Triple point of water on absolute scale $\\$ $=273.16*\dfrac{9}{5}=491.69 K$

5   Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:$\\$ $\begin{array}{|c|c|c|c|} \hline \textbf{Temperature} & \textbf{Pressure thermometer A} &\textbf{Pressure thermometer B} \\ \hline \text{Triple-point of water } &1.250 * 10^ 5 \ \ Pa & 0.200 *10^5 \ \ Pa \\ \hline \text{ Normal melting point of sulphur} &1.797 * 10^5 \ \ Pa & 0.287 * 10^5 \ \ Pa \\ \hline \end{array}$$\\$ (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?

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:$\\$ $ \dfrac{P_A}{T}=\dfrac{P_1}{T_1}\\ \therefore T_1=\dfrac{P_1T}{P_A}=\dfrac{1.797*10^5*273.16}{1.25*10^5}\\ =392.69K$$\\$ Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer $A$ is $392.69 K.$$\\$ 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:$\\$ $\dfrac{P_B}{T}=\dfrac{P_1}{T_1}\\ \dfrac{0.200*10^5}{273.16} =\dfrac{0.287*10^5}{T_1}\\ \therefore T_1=\dfrac{0.287*10^5}{0.200*10^5}*273.16\\ =391.98K$$\\$ Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer $B$ is $391.98 K.$$\\$ (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.