JEE Advanced 2023 — Paper 1 — CHEM-III solution

JEE Advanced2023Paper 1CHEM-III

One mole of an ideal monoatomic gas undergoes two reversible processes ( $A \to B$ and $B \to C$ ) as shown in the given figure:

$A \to B$ is an adiabatic process. If the total heat absorbed in the entire process ( $A \to B$ and $B \to C$ ) is $R T_{2} ln 10$ , the value of $2 lo g V_{3}$ is _________.

[Use, molar heat capacity of the gas at constant pressure, $C_{p, m} = \frac{5}{2} R$ ]

Your answer

Solution

For A $\to$ B

(Reversible adiabatic)

$600V_1^{\gamma-1}=60V_2^{\gamma-1} \qquad \left(\gamma=\dfrac{5}{3}\right)$

$\Rightarrow 600 (V_{1})^{2/3} = 60 (V_{2})^{2/3}$

$\Rightarrow 10 = (\frac{V _{2}}{V _{1}})^{2/3}$

$\Rightarrow 10 = (\frac{V _{2}}{10})^{2/3}$

$\Rightarrow V_{2} = 10 (10)^{3/2} = 1 0^{5/2}$

Now,

$q_{\text{net}}=RT_2\ln 10 =60R\ln 10 =q_{AB}+q_{BC}$

$∵ q_{A B} = 0$

$\Rightarrow q_{BC}=60R\ln 10 =60R\ln\left(\dfrac{V_3}{V_2}\right)$

[ $∵$ B $\to$ C is reversible isothermal]

$\Rightarrow 60R\ln 10 =60R\ln\left(\dfrac{V_3}{10^{5/2}}\right)$

$\Rightarrow \log 10 =\log V_3-\dfrac{5}{2}$

$\Rightarrow lo g V_{3} = \frac{7}{2}$

$\Rightarrow 2 lo g V_{3} = 7$

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