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Heat and Thermodynamics question

2022 · 30 Jun · Shift 1 · Q49
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  5. /2022 · 30 Jun · Shift 1 · Q49

Heat and Thermodynamics question

2022 · 30 Jun · Shift 1 · Q49

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The pressure of the gas in a constant volume gas thermometer is 100 cm of mercury when placed in melting ice at 1 atm. When the bulb is placed in a liquid, the pressure becomes 180 cm of mercury. Temperature of the liquid is : (Given 0 ∘^\circ∘ C = 273 K)
  1. A
    300 K
  2. B
    400 K
  3. C
    600 K
  4. D
    491 K
View written solutionFree

Correct answer: D

  1. Use the principle of a constant volume gas thermometer

At constant volume, the pressure of a gas is directly proportional to its absolute temperature:

PT=constant\frac{P}{T}=\text{constant}TP​=constant

So,

P1T1=P2T2\frac{P_1}{T_1}=\frac{P_2}{T_2}T1​P1​​=T2​P2​​

  1. Identify the given values

  • In melting ice: T1=273 KT_1=273\,\text{K}T1​=273K P1=100 cm of HgP_1=100\,\text{cm of Hg}P1​=100cm of Hg

  • In the liquid: P2=180 cm of HgP_2=180\,\text{cm of Hg}P2​=180cm of Hg T2=?T_2=?T2​=?

  1. Apply the relation

100273=180T2\frac{100}{273}=\frac{180}{T_2}273100​=T2​180​

Therefore,

T2=273×180100T_2=273\times \frac{180}{100}T2​=273×100180​

T2=273×1.8=491.4 KT_2=273\times 1.8=491.4\,\text{K}T2​=273×1.8=491.4K

  1. Choose the closest option

T2≈491 KT_2\approx 491\,\text{K}T2​≈491K

So the correct option is D.

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