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INTRODUCTION - Chapter 11 - Thermal Properties Of Fluids - Ncert Solutions class 11 - Physics


Question 11.1:

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.

Answer:

Kelvin and Celsius scales are related as:

TC = TK – 273.15 … (i)

Celsius and Fahrenheit scales are related as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/189/4645/NS_17-10-2008_Sravana_11_Physics_11_22_NRJ_LVN_html_2f2e9445.gif  … (ii)

For neon:

TK = 24.57 K

TC = 24.57 – 273.15 = –248.58°C

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/189/4645/NS_17-10-2008_Sravana_11_Physics_11_22_NRJ_LVN_html_2f17e016.gif

For carbon dioxide:

TK = 216.55 K

TC= 216.55 – 273.15 = –56.60°C

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/189/4645/NS_17-10-2008_Sravana_11_Physics_11_22_NRJ_LVN_html_me6bbc13.gif

Question 11.2:

Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB?

Answer:

Triple point of water on absolute scaleA, T1 = 200 A

Triple point of water on absolute scale B, T2 = 350 B

Triple point of water on Kelvin scale, TK = 273.15 K

The temperature 273.15 K on Kelvin scale is equivalent to 200 A on absolute scale A.

T1 = TK

200 A = 273.15 K

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/189/4646/NS_17-10-2008_Sravana_11_Physics_11_22_NRJ_LVN_html_m75b8961d.gif

The temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B.

T2 = TK

350 B = 273.15

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/189/4646/NS_17-10-2008_Sravana_11_Physics_11_22_NRJ_LVN_html_346cb82f.gif

TA is triple point of water on scale A.

TB is triple point of water on scale B.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/189/4646/NS_17-10-2008_Sravana_11_Physics_11_22_NRJ_LVN_html_3fe713a.gif

Therefore, the ratio TA : Tis given as 4 : 7.

Question 11.3:

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:

Ro [1 + α (– To)]

The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?

Answer:

It is given that:

R0 [1 + α (– T0)] … (i)

Where,

R0 and T0 are the initial resistance and temperature respectively

R and T are the final resistance and temperature respectively

α is a constant

At the triple point of water, T0 = 273.15 K

Resistance of lead, R0 = 101.6 Ω

At normal melting point of lead, T = 600.5 K

Resistance of lead, R = 165.5 Ω

Substituting these values in equation (i), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/189/8090/Chapter%2011_html_m3102532e.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/189/8090/Chapter%2011_html_m28fe770e.gif

For resistance, R1 = 123.4 Ω

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/189/8090/Chapter%2011_html_m68987568.gif

Question 11.4:

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 °C and 100 °C 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?

(c) The absolute temperature (Kelvin scale) is related to the temperature tc on the Celsius scale by

tc = – 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?

Answer:

(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°C 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 tc, on Celsius scale as:

tc = T – 273.15

(d) Let TF be the temperature on Fahrenheit scale and TK be the temperature on absolute scale. Both the temperatures can be related as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/189/4648/NS_17-10-2008_Sravana_11_Physics_11_22_NRJ_LVN_html_664ef7bd.gif

Let TF1 be the temperature on Fahrenheit scale and TK1 be the temperature on absolute scale. Both the temperatures can be related as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/189/4648/NS_17-10-2008_Sravana_11_Physics_11_22_NRJ_LVN_html_16327c86.gif

It is given that:

TK1 – TK = 1 K

Subtracting equation (i) from equation (ii), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/189/4648/NS_17-10-2008_Sravana_11_Physics_11_22_NRJ_LVN_html_3deb6750.gif

Triple point of water = 273.16 K

∴Triple point of water on absolute scale = https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/189/4648/NS_17-10-2008_Sravana_11_Physics_11_22_NRJ_LVN_html_m207305d7.gif  = 491.69