Thermal Properties of Matter

11.5 (a) For Thermometer A

Triple point of water, T = 273.16 K

At this temperature, the pressure in thermometer A , $P_A$ = 1.250 $*{10}^5$ Pa

Let $T_1$ be the temperature for the normal melting point of sulphur and $P_1$ be the corresponding pressure. It is given, $P_1$ = 1.797 $*{10}^5$ Pa

From Charles' law, we get $\frac{P_A}{T}$ = $\frac{P_1}{T_1}$ , $T_1$ = $\frac{P_1*\mathrm{T}}{P_A}$ = $\frac{1.797\mathrm{}*{10}^5*273.16}{1.250\mathrm{}*{10}^5}$ = 392.69 K

For Thermometer B

Triple point of water, T = 273.16 K

At this temperature, the pressure in thermometer B , $P_B$ = 0.2 $*{10}^5$ Pa

Let $T_1$ be the temperature for the normal melting point of sulphur and $P_1$ be the corresponding pressure. It is given, $P_1$ = 0.287 $*{10}^5$ Pa

From Charles' law, we get $\frac{P_B}{T}$ = $\frac{P_1}{T_1}$ , $T_1$ = $\frac{P_1*\mathrm{T}}{P_B}$ = $\frac{0.287\mathrm{}*{10}^5*273.16}{0.2\mathrm{}*{10}^5}$ = 391.98 K

(b) The oxygen and hydrogen gas present in thermometers A and B respectively are not perfect ideal gas, hence there is a slight difference in readings.

(c) To reduce the difference between two readings, the measurement should be carried out at low pressure so that gases behave like an ideal gas.

11.4 (a) The triple point of water has a unique value of 273.16 K, irrespective of pressure and volume. Whereas, melting point of ice and boiling point of water, the temperature value depends on pressure and volume.

(b) The other fixed point on Kelvin scale is 0 K.

(c) The temperature 273.16 K is the triple point of water, it is not the melting point of ice. The melting point of ice is specified in Celsius scale as 0 $°C.$ Hence the absolute temperature in Kelvin scale, $T_k$ is related to temperature $T_c$ in Celsius scale as

$T_c=T_k-273.15$

(d) Let $T_f$ and $T_k$ be the temperature in Fahrenheit and absolute scale. From the co-relation of the scale we know

$\frac{T_f-32}{9}$ = $\frac{T_k-273.15}{5}$ ……(i)

Let $T_{f1}$ and $T_{k1}$ be the temperature in Fahrenheit and absolute scale. From the co-relation of the scale we know

$\frac{T_{f1}-32}{9}$ = $\frac{T_{k1}-273.15}{5}$ ……(ii)

It is given $T_{k1}-T_k$ = 1 K

Now subtracting (i) from (ii), we get

$\frac{T_{f1}-32}{9}$ - $\frac{T_f-32}{9}$ = $\frac{T_{k1}-273.15}{5}-\frac{T_k-273.15}{5}$

$\frac{T_{f1}-T_f}{9}$ = $\frac{T_{k1}-T_k}{5}$

$T_{f1}-T_f$ = $\frac{9}{5}$

Triple point of water = 273.16 K

Hence triple point of water in the absolute scale = 273.16 $*$ $\frac{9}{5}$ = 491.69