Kinetic Theory of Gases and ldeal Gas | Physics JEE previous year questions with Solutions

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Topic Question Set

Q 1 :
Water of mass m gram is slowly heated to increase the temperature from $T_{1}$ to $T_{2}$ . The change in entropy of the water, given specific heat of water is $1 {Jkg}^{- 1} K^{- 1}$ , is [2025]

zero
$m (T_{2} - T_{1})$
$m \ln (\frac{T_{1}}{T_{2}})$
$m \ln (\frac{T_{2}}{T_{1}})$

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(4)

dQ = msdT

$d S = \frac{d Q}{T} = \frac{m s d T}{T}$

$△ S = \int \frac{m s d T}{T} = m s \ln \frac{T_{f}}{T_{i}}$

$△ S = m s \ln \frac{T_{2}}{T_{1}} as s = 1$

Q 2 :
A Carnot engine (E) is working between two temperatures 473 K and 273 K. In a new system two engines - engine $E_{1}$ works between 473 K to 373 K and engine $E_{2}$ works between 373 K to 273 K. If $η_{12}$ , $η_{1}$ and $η_{2}$ are the efficiencies of the engines E, $E_{1}$ and $E_{2}$ , respectively, then [2025]

$η_{12} < η_{1} + η_{2}$
$η_{12} = η_{1} η_{2}$
$η_{12} = η_{1} + η_{2}$
$η_{12} \geq η_{1} + η_{2}$

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(1)

Efficiencies of a Carnot engine $η = 1 - \frac{T_{sink}}{T_{source}}$

$\Rightarrow η_{1} = 1 - \frac{373 K}{473 K} = \frac{100}{473}$

$η_{2} = 1 - \frac{273 K}{373 K} = \frac{100}{373}$

$η_{12} = 1 - \frac{273 K}{473 K} = \frac{100}{473}$

$η_{12} - η_{1} = \frac{200}{473} - \frac{100}{473} = \frac{100}{473} < \frac{100}{373}$

$\Rightarrow η_{12} - η_{1} < η_{2}$

or $η_{12} < η_{1} + η_{2}$

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