JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Chemical Thermodynamics

An ideal gas,

\overline{C}_{V} = \frac{5}{2} R

, is expanded adiabatically against a constant pressure of 1 atm until it doubles in volume. If the initial temperature and pressure is 298 K and 5 atm, respectively then the final temperature is ______ K (nearest integer).
[

\overline{C}_{V}

is the molar heat capacity at constant volume]

Answer: 274

Solution

Adiabatic (

q = 0

) expansion against constant external pressure:

Δ U = w

, i.e.

n C_{V} (T_{2} - T_{1}) = - P_{e x t} (V_{2} - V_{1})

with

V_{2} = 2 V_{1}

.
Using

V_{1} = \frac{n R T _{1}}{P _{1}}

and

P_{e x t} = 1

atm,

P_{1} = 5

atm,

T_{1} = 298

C_{V} = \frac{5}{2} R

\frac{5}{2} R (T_{2} - 298) = - 1 \cdot (2 \cdot \frac{R \cdot 298}{5} - \frac{R \cdot 298}{5}) = - \frac{R \cdot 298}{5} .

Solving gives

T_{2} = 274.16 \approx 274

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Integer answer

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JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Chemical Thermodynamics

Solution

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