NEET-XII-Physics

23: Heat and Temperature

with Solutions - page 3
Qstn# iv-4 Prvs-QstnNext-Qstn
  • #4
    The pressure measured by a constant volume gas thermometer is 40 kPa at the triple point of water. What will be the pressure measured at the boiling point of water (100°C)?
    Ans : Given:
    Pressure measured by a constant volume gas thermometer at the triple point of water, Ptr = 40 kPa = 40 × 103 Pa
    Boiling point of water, T = 100°C = 373.16 K
    Let the pressure measured at the boiling point of water be P.
    ​For a constant volume gas thermometer, temperature-pressure relation is given below:
    `` T=\frac{P}{{P}_{tr}}\times 273.16\,\mathrm{\,K\,}``
    `` \Rightarrow P=\frac{T\times {P}_{tr}}{273.16}``
    `` \Rightarrow P=\frac{373.16\times 40\times {10}^{3}}{273.16}``
    `` \Rightarrow P=54643\,\mathrm{\,Pa\,}``
    `` \Rightarrow P=54.6\times {10}^{3}\,\mathrm{\,Pa\,}``
    `` \Rightarrow P\simeq 55\,\mathrm{\,kPa\,}``
    Therefore, the pressure measured at the boiling point of water is 55 kPa.
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