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Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IFirst law of ThermodynamicsEasy2 minPYQ_2020

ChemistryEasysingle choice

One mole of an ideal gas $(C_{V} = 20 J K^{- 1} m o l^{- 1})$ initially at STP is heated at constant volume to twice the initial temperature. For the process, W and q will be-

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Answer:

Solution:

$T_{1} = 273 K$
$T_{2} = 2 \times 273 K$
$\Rightarrow ∆ T = 273$
Volume is constant
So, $∆ V = 0$
$W = P Δ V = P \times 0 = 0$
$q = n C_{V} (Δ T) = 20 \times 273 = 5460 J = 5.46 k J$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Thermodynamics - ISubtopic:First law of Thermodynamics

⏱ 2mℹ️ Source: PYQ_2020

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