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 point of sulphur |
1.797 × 10^{5}Pa |
0.287 × 10^{5} Pa |

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

**(b)** What do you think is the reason behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

Asked by Pragya Singh | 1 year ago | 78

**(a)** Triple point of water, T= 273.16K.

Pressure in thermometer A at the triple point, P_{A} =1.25×10^{5} Pa

Normal melting point of sulphur = T_{1}

Pressure in thermometer A at this temperature, P_{1}=1.797×10^{5} Pa

According to Charles law, we have the relation:

\( \dfrac{P_A}{T}=\dfrac{P_1}{T_1}\)

\(T_1= \dfrac{P_1 T}{P_A}\)

= \( \dfrac{(1.797×10^5×273.16)}{1.25×10^5}\)

=392.69 K

Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer A is 392.69 K.

**(b)** The pressure in thermometer B at the triple point of water, P_{B} =0.2×10^{5 }Pa

The temperature in thermometer B at the normal melting point (T_{1}) of sulphur is,

P_{1} = 0.287×10^{5} Pa

According to Charles law, we can write the relation:

\( \dfrac{P_B}{T}=\dfrac{P_1}{T_1}\)

\( T_1= \dfrac{P_1 T}{P_B}\)

T_{1 }= \(
\dfrac{ (0.287×10^5 × 273.16)}{0.2×10^5}\)

=391.98 K

Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer B is 391.98 K.

**(b)** The reason for the discrepancy is that the gases do not behave like ideal gas in practice. The discrepancy can be reduced by taking the reading at very low pressure so that the gases show perfect behaviour.

A hot ball cools from 90°C to 10°C in 5 minutes. If the surrounding temperature is 20°C, what is the time taken to cool from 60°C to 30°C?

Answer the following questions based on the P-T phase diagram of carbon dioxide:

**(a)** At what temperature and pressure can the solid, liquid and vapour phases of \( CO_2\) co-exist in equilibrium?

**(b)** What is the effect of the decrease of pressure on the fusion and boiling point of \( CO_2\)?

**(c) **What are the critical temperature and pressure for \( CO_2\)? What is its significance?

**(d)** Is \( CO_2\) solid, liquid or gas at

**(a) **–70°C under 1 atm,

**(b)** –60°C under 10 atm,

**(c) **15°C under 56 atm?

A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C.

Explain why :

**(a)** a body with large reflectivity is a poor emitter

**(b)** a brass tumbler feels much colder than a wooden tray on a chilly day

**(c)** an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace

**(d) **the earth without its atmosphere would be inhospitably cold

**(e)** heating systems based on the circulation of steam are more efficient in warming a building than those based on the circulation of hot water

A brass boiler has a base area of 0.15 m^{2} and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s^{–1} m^{–1 }K^{–1}; Heat of vaporisation of water = 2256 × 10^{3} J kg^{–1}.