One mole of an ideal monoatomic gas undergoes two reversible processes (A → B and B → C) as shown in the given figure: (see figure) \(\mathrm{A} \rightarrow \mathrm{B}\) is an adiabatic process. If the total heat absorbed in the entire process (\(\mathrm{A} \rightarrow \mathrm{B}\) and \(\mathrm{B} \rightarrow \mathrm{C}\)) is \(\mathrm{RT}_{2} \ln 10\), the value of \(2 \log \mathrm{~V}_{3}\) is ____. [Use, molar heat capacity of the gas at constant pressure, \(\mathrm{C}_{\mathrm{p}, \mathrm{m}}=\frac{5}{2} \mathrm{R}\)]

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Chemical Thermodynamics

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

(see 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

RT_{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

]

Answer: 7

For the adiabatic process A → B with

γ = 5/3

for monoatomic gas:

T_{A} V_{A}^{γ - 1} = T_{B} V_{B}^{γ - 1}

600 V_{1}^{2/3} = 60 V_{2}^{2/3}

10 = (V_{2} / V_{1})^{2/3}

With

V_{1} = 10

V_{2} = 1 0^{5/2}

Total heat absorbed:

q_{n e t} = q_{A B} + q_{B C} = 0 + q_{B C} = R T_{2} ln 10

For the isothermal process B → C at temperature

T_{2} = 60

q_{B C} = n R T_{2} ln \frac{V _{3}}{V _{2}} = 60 R ln \frac{V _{3}}{1 0 ^{5/2}}

Equating:

60 R ln 10 = 60 R ln \frac{V _{3}}{1 0 ^{5/2}}

ln 10 = ln V_{3} - \frac{5}{2} ln 10

ln V_{3} = \frac{7}{2} ln 10

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

Therefore:

2 lo g V_{3} = 7

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Integer answer