1. Tardigrade
  2. Question
  3. Chemistry
  4. 1 Mole of CO2 gas at 300 K expanded under the reversible adiabatic condition such that its volume becomes 27 times. The magnitude of work done (i n k J / m o l) is: (Given textγ = 1.33 and Cv=25.10Jmol- 1K- 1forCO2 ) report your answer by rounding it up to nearest whole number

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Q. $1$ Mole of $CO_{2}$ gas at $300 \, K$ expanded under the reversible adiabatic condition such that its volume becomes $27$ times. The magnitude of work done $\left(i n k J / m o l\right)$ is:
(Given $\text{γ} = 1.33$ and $ \, C_{v}=25.10Jmol^{- 1}K^{- 1}forCO_{2}$ )
report your answer by rounding it up to nearest whole number

NTA AbhyasNTA Abhyas 2022

Solution:

Number of moles $=1$
$T_{1}=300K$
$V_{2}=27V_{1}$
$TV_{1}^{\gamma - 1}=T_{2}V_{2}^{\gamma - 1}$
$\left(\frac{T_{1}}{T_{2}}\right)=\left(\frac{V_{2}}{V_{1}}\right)^{\gamma - 1}$
$T_{2}=300\left(\frac{1}{27}\right)^{\frac{1}{3}}$
$T_{2}=100K$
Adiabatic condition, $q=0;\Delta E=w=nC_{v}\left(T_{2} - T_{1}\right)$
$w=1\times 25.10\left(100 - 300\right)=-5020J / m o l e$
$w=-5.02k J / m o l e$