CBSE Class 11 Physics Chapter 11: Thermal Properties of Matter NCERT Solutions

We are given the triple points of neon and carbon dioxide in Kelvin and need to convert them to Celsius and Fahrenheit. The relationships between the scales are:

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

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

For Neon:

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

Using Equation (i) to convert to Celsius:

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

Now, using Equation (ii) to convert this Celsius temperature to Fahrenheit:

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

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

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

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

For Carbon Dioxide:

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

Using Equation (i) to convert to Celsius:

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

Now, using Equation (ii) to convert this Celsius temperature to Fahrenheit:

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

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

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

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 T_{\rm K} = 273.15 \text{ K}.

On scale A, the triple point of water is T_A = 200 \text{ A}. This means that 200 units on scale A correspond to 273.15 K.

Therefore, the conversion factor for scale A is: 1 \text{ A} = \frac{273.15}{200} \text{ K}.

On scale B, the triple point of water is T_B = 350 \text{ B}. This means that 350 units on scale B also correspond to 273.15 K.

Therefore, the conversion factor for scale B is: 1 \text{ B} = \frac{273.15}{350} \text{ K}.

We want to find the relation between a temperature T_A on scale A and T_B on scale B. Since both scales are absolute and calibrated using the triple point of water, we can set up a proportion based on the Kelvin scale equivalence.

Let T_A be a temperature on scale A and T_B be the equivalent temperature on scale B. We know that:

T_A \text{ A} = T_{\rm K} and T_B \text{ B} = T_{\rm K}

This implies T_A \text{ A} = T_B \text{ B}.

Substituting the Kelvin equivalents:

T_A \left( \frac{273.15}{200} \right) \text{ K} = T_B \left( \frac{273.15}{350} \right) \text{ K}

We can cancel out the 273.15 from both sides:

\frac{T_A}{200} = \frac{T_B}{350}

This equation gives the relationship between temperatures on scales A and B. We can rearrange it to express T_A in terms of T_B or vice versa:

T_A = \frac{200}{350} T_B = \frac{4}{7} T_B Or, T_B = \frac{350}{200} T_A = \frac{7}{4} T_A

Thus, the relation between T_A and T_B is \frac{T_A}{200} = \frac{T_B}{350}.