Q:1. The triple points of Neon and Carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures in Celsius and Fahrenheit scales.

Sol. If temperature in Kelvin scale is K and that in Celsius scale is C then, C = K – 273.15

For Neon, C = 24.57 – 273.15 = -248.58 ^{o} C

For CO2, C = 216.55 – 273.15 = -56.60 ^{o}C

If temperature in Fahrenheit scale is F then,

$ \displaystyle \frac{F-32}{180} = \frac{K-273.15}{100} $

For Neon, $ \displaystyle F = \frac{180}{100}(24.57-273.15) + 32 $

= – 415.44^{o}F

For CO_{2} , $ \displaystyle F = \frac{180}{100}(216.55-273.15) + 32 $

= -69.88 °F

Q: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} .

Sol. According to the question,

Triple point of water = 200 A = 350 B = 273.16 K

[Triple point of water in Kelvin scale = 273.16 K]

$ \displaystyle 1A = \frac{273.16}{200}K $

$ \displaystyle 1B = \frac{273.16}{350}K $

If T_{A} and T_{B} represent the triple point of water on scales A and B then,

$ \displaystyle \frac{273.16}{200}T_A = \frac{273.16}{350}T_B $

$ \displaystyle \frac{T_A}{T_B} = \frac{200}{350}=\frac{4}{7} $

$ \displaystyle T_A = \frac{4}{7}T_B $

Q:3. The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law : R = R_{0} [1 + α(T – T_{0})].

The resistance is 10.1 16 ohm at the triple pint of water 273.16 K, and 165.5 ohm at the normal melting point of lead 600.5 K. What is the temperature when the resistance is 123.4 ohm

Sol. R_{0} = 101.6 ohm, T_{0} = 273.16 K,

R_{1} = 165.5 ohm, T_{1} = 600.5 K,

R_{2} = 123.4 ohm , T_{2} = ?

Using the relation, R = R_{0} [1 + α (T – T_{0})]

165.5 = 101.6 [1 + α (600.5 – 273.16)]

= 101.6[1 + α × 327.34]

$ \displaystyle \alpha = \frac{165.5-101.6}{101.6 \times 327.34} $

Also, 123.4 = 101.6 [1 + α(T_{2} – 273.16)]

Substituting the value of α in the equation

T_{2} = 384.83 K

Q:4. Answer the following:

(a)The triple point of water is a standard fixed point in modern thermometry Why ? . What is wrong 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) T is related to temperature tc on the Celsius scale by t_{c} = T – 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 ?

Sol.

(a) Triple point of water has a unique value (273.16 K) at fixed values of pressure and volume. But melting point of ice and boiling point of water do not have unique value as they depend both on pressure and temperature.

(b)The other fixed point is the absolute zero.

(c) On Celsius scale 0°C corresponds to melting point of ice at normal pressure and the same temperature in Kelvin scale is 273.15 K. The triple point of water in Kelvin scale is 273.16 K.

By the relation, tc = T – 273.15,

The triple point of water on Celsius scale = 273.16 – 273.15 = 0.01 °C

(d) Relation between temperature in Fahrenheit scale and absolute scale is

(F – 32)/180 = (K – 273.15)/100 …… (i)

For another set of temperature,

(F’ – 32)/180 = (K’-273.15)/100 …… (ii)

Subtracting (i) from (ii), we get

(F’ – F)/180 = (K’-K)/100

F’ – F = (180/100) × (K’ – K)

If K’ – K = 1 K,

then F’ – F = (180/100) × 1 = 9/5

The triple point of water (273.16 K) in the new scale is 273.16 × (9/5) = 491.69

Q:5. Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:

Temperature | Pressure thermometer A | Pressure thermometer B |

Triple point of water | 1.250 × 10^{5} Pa |
0.200 × 10^{5} Pa |

Normal melting | 1.797 × 10^{5} Pa |
0.287 × 10^{5} Pa |

(a)What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?

(b)What do you think about the reason of the slightly different answer from A and B? (The thermometers are to tally). What further procedure is needed in the experiment to reduce the discrepancy between the two readings.

Sol: Triple point of water = 273.16 K, Let T be the melting point of sulphur.

For thermometer A, $\displaystyle T = \frac{P}{P_w}\times 273.16 $

$\displaystyle T = \frac{1.797\times 10^5}{1.250\times 10^5}\times 273.16 $

= 392.69 K [∵ P∝ T]

For thermometer B,$ \displaystyle T = \frac{0.287\times 10^5}{0.200\times 10^5}\times 273.16 $

T = 391.98 K

The cause of this difference in the two temperatures is that oxygen and hydrogen gases are not perfectly ideal.

To reduce the discrepancy, the readings should be taken at a very low pressure where the gases approach to the ideal gas behavior.

Q:6. A steel tape 1 m long is correctly calibrated for a temperature of 27 °C. the length of a steel rod measured by this tape is found to be 63 cm on a hot day when the temperature is 45 °C. What is the actual length of the steel rod on that day ? what is the length of the same steel rod on a day when the temperature is 27 °C? Cefficient of linear expansion of steel = 1.2 × 10^{-5}/0C

Sol. L_{1} = 100 cm & T_{1} = 27 °C

At 45 °C , L_{2} = L_{1} + α L1∆T

= 100 + 1.2 × 10-5 × 100 × (45 – 27) = 100.0216 cm

Length of 1 cm mark at 27 °C on this scale becomes at 45 °C = 100.0216/100 cm.

Actual length of the steel rod on that day = (100.0216/100 )× 63 = 36.0136 cm

Also, length of the same rod at 27 °C = 63 × 1 = 63 cm