# 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)]$$\\ Where, T is the temperature when the resistance of lead is 123.4\Omega$$\\$ $123.4 = 101.6 [ 1 + 1.92 * 10^{-3} ( T - 273.15)]\\ 1.214 = 1 + 1.92 * 10^{-3} (T - 273.15)\\ \dfrac{0.214}{1.92*10^{-3}} \\ =T-273.15\\ \therefore T=384.61K$

##### 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.