CBSE Class 11 Physics NCERT Solutions: Thermal Properties of Matter

We are given the triple points of neon and carbon dioxide in Kelvin and need to convert them to Celsius and Fahrenheit. We use the following standard conversion formulas:

1. Kelvin to Celsius: $$T\_\{\backslash rm\; C\}\; =\; T\_\{\backslash rm\; K\}\; -\; 273.15$$

2. Celsius to Fahrenheit: $$T\_\{\backslash rm\; F\}\; =\; \backslash frac\{9\}\{5\}T\_\{\backslash rm\; C\}\; +\; 32$$

For Neon:

Given temperature in Kelvin, $T\_\{\backslash rm\; K\}\; =\; 24.57\; \backslash \; \{\backslash rm\; K\}$.

Converting to Celsius:

$T\_\{\backslash rm\; C\}\; =\; 24.57\; -\; 273.15\; =\; -248.58^\{\backslash circ\}\{\backslash rm\; C\}$

Converting this Celsius temperature to Fahrenheit:

$T\_\{\backslash rm\; F\}\; =\; \backslash frac\{9\}\{5\}(-248.58)\; +\; 32$

$T\_\{\backslash rm\; F\}\; =\; -447.444\; +\; 32$

$T\_\{\backslash rm\; F\}\; =\; -415.44^\{\backslash circ\}\{\backslash rm\; F\}$

So, the triple point of neon is -248.58°C and -415.44°F.

For Carbon Dioxide:

Given temperature in Kelvin, $T\_\{\backslash rm\; K\}\; =\; 216.55\; \backslash \; K$.

Converting to Celsius:

$T\_\{\backslash rm\; C\}\; =\; 216.55\; -\; 273.15\; =\; -56.60^\{\backslash circ\}\{\backslash rm\; C\}$

Converting this Celsius temperature to Fahrenheit:

$T\_\{\backslash rm\; F\}\; =\; \backslash frac\{9\}\{5\}(-56.60)\; +\; 32$

$T\_\{\backslash rm\; F\}\; =\; -101.88\; +\; 32$

$T\_\{\backslash rm\; F\}\; =\; -69.88^\{\backslash circ\}\{\backslash rm\; F\}$

So, the triple point of carbon dioxide is -56.60°C and -69.88°F.

We are given two absolute temperature scales, A and B. The triple point of water is defined as 200 units on scale A and 350 units on scale B. The standard value for the triple point of water on the Kelvin scale is 273.15 K.

Since both scales are absolute, their zero points correspond to absolute zero. The relationship between any two absolute scales can be established by comparing their readings to a known fixed point, such as the triple point of water.

Let $T\_A$ be the temperature on scale A and $T\_B$ be the temperature on scale B. We know that:

Triple point of water on scale A = 200 A

Triple point of water on scale B = 350 B

Triple point of water on Kelvin scale = 273.15 K

For scale A, the temperature $T\_A$ corresponds to the Kelvin temperature $T\_K$ such that:

$\backslash frac\{T\_A\}\{200\; \backslash text\{\; A\}\}\; =\; \backslash frac\{T\_K\}\{273.15\; \backslash text\{\; K\}\}$

For scale B, the temperature $T\_B$ corresponds to the Kelvin temperature $T\_K$ such that:

$\backslash frac\{T\_B\}\{350\; \backslash text\{\; B\}\}\; =\; \backslash frac\{T\_K\}\{273.15\; \backslash text\{\; K\}\}$

Since both expressions are equal to $\backslash frac\{T\_K\}\{273.15\; \backslash text\{\; K\}\}$, we can equate them to find the relationship between $T\_A$ and $T\_B$:

$$\backslash frac\{T\_A\}\{200\}\; =\; \backslash frac\{T\_B\}\{350\}$$

This equation shows the direct relationship between the temperatures measured on absolute scales A and B, based on their respective definitions of the triple point of water.