Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IFirst law of ThermodynamicsMedium2 minPYQ_2019

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Under the isothermal condition, a gas at $300 K$ expands from $0.1 L$ to $0.25 L$ against a constant external pressure of $2$ bar. The work done by the gas is

(Given that $1 L$ bar $= 100 J$ )

NEET 2019 (All India)

Options:

Answer:

Solution:

The work done by a gas,

$W = P_{external} \times ΔV$

$= - 2 \times (v_{2} - v_{1})$

$= - 2 \times (0.25 - 0.10)$

$= - 0.30 L$ bar

Given, $1 L$ bar $= 100 J$ .

So, $W = - 0 .3 \times 100 = - 30 J$ .

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

⏱ 2mℹ️ Source: PYQ_2019

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