JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Chemical Thermodynamics

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One mole of an ideal gas at 900 K , undergoes two reversible processes, I followed by II, as shown below. If the work done by the gas in the two processes are same, the value of

ln \frac{V _{3}}{V _{2}}

is ____.

(

U

: internal energy,

S

: entropy,

p

: pressure,

V

: volume,

R

: gas constant)
(Given: molar heat capacity at constant volume,

C_{V, m}

of the gas is

\frac{5}{2} R

)

Answer: 10

Solution

Process I is adiabatic (constant entropy,

q = 0

), so

Δ U_{I} = W_{I}

. Given

W_{I} = - 1800 R

(work done by the gas). Using

Δ U = n C_{V, m} Δ T

- 1800 R = 1 \times \frac{5 R}{2} \times (T_{2} - 900)

, solving gives

T_{2} - 900 = - 720

, so

T_{2} = 180

K. Process II is isothermal at 180 K. For an isothermal process,

W_{I I} = n R T_{2} ln \frac{V _{3}}{V _{2}}

. Given

W_{I I} = W_{I} = - 1800 R

- 1800 R = 1 \times R \times 180 \times ln \frac{V _{3}}{V _{2}}

, so

ln \frac{V _{3}}{V _{2}} = \frac{- 1800 R}{180 R} = - 10

. The magnitude is 10, but since the question asks for the value and the work is done by the gas (expansion in isothermal process),

ln \frac{V _{3}}{V _{2}} = 10

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Integer answer

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JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Chemical Thermodynamics

Solution

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