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physics ncert solutions class 11th Physics Class 11th Thermal Properties of Matter

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

Temperature                         Pressure                     Pressure

thermometer A thermometer B

Triple-point of water            1.250 × 10⁵ Pa            0.200 × 10⁵ Pa

Normal melting point           1.797 × 10⁵ Pa            0.287 × 10⁵ Pa

of sulphur

(a) What is the absolute temperature of 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?

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Payal Gupta | Contributor-Level 10

4 months ago

11.5 (a) For Thermometer A

Triple point of water, T = 273.16 K

At this temperature, the pressure in thermometer A , $P_{A}$ = 1.250 $* 10^{5}$ Pa

Let $T_{1}$ be the temperature for the normal melting point of sulphur and $P_{1}$ be the corresponding pressure. It is given, $P_{1}$ = 1.797 $* 10^{5}$ Pa

From Charles' law, we get $\frac{P_{A}}{T}$ = $\frac{P_{1}}{T_{1}}$ , $T_{1}$ = $\frac{P_{1} * T}{P_{A}}$ = $\frac{1.797 * 10^{5} * 273.16}{1.250 * 10^{5}}$ = 392.69 K

For Thermometer B

Triple point of water, T = 273.16 K

At this temperature, the pressure in thermometer B , $P_{B}$ = 0.2 $* 10^{5}$ Pa

Let $T_{1}$ be the temperature for the normal melting point of sulphur and $P_{1}$ be the co

...more

11.5 (a) For Thermometer A

Triple point of water, T = 273.16 K

At this temperature, the pressure in thermometer A , $P_{A}$ = 1.250 $* 10^{5}$ Pa

Let $T_{1}$ be the temperature for the normal melting point of sulphur and $P_{1}$ be the corresponding pressure. It is given, $P_{1}$ = 1.797 $* 10^{5}$ Pa

From Charles' law, we get $\frac{P_{A}}{T}$ = $\frac{P_{1}}{T_{1}}$ , $T_{1}$ = $\frac{P_{1} * T}{P_{A}}$ = $\frac{1.797 * 10^{5} * 273.16}{1.250 * 10^{5}}$ = 392.69 K

For Thermometer B

Triple point of water, T = 273.16 K

At this temperature, the pressure in thermometer B , $P_{B}$ = 0.2 $* 10^{5}$ Pa

Let $T_{1}$ be the temperature for the normal melting point of sulphur and $P_{1}$ be the corresponding pressure. It is given, $P_{1}$ = 0.287 $* 10^{5}$ Pa

From Charles' law, we get $\frac{P_{B}}{T}$ = $\frac{P_{1}}{T_{1}}$ , $T_{1}$ = $\frac{P_{1} * T}{P_{B}}$ = $\frac{0.287 * 10^{5} * 273.16}{0.2 * 10^{5}}$ = 391.98 K

(b) The oxygen and hydrogen gas present in thermometers A and B respectively are not perfect ideal gas, hence there is a slight difference in readings.

(c) To reduce the difference between two readings, the measurement should be carried out at low pressure so that gases behave like an ideal gas.

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