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Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IAdiabatic Process/PolytropicMedium2 minQB

ChemistryMediumnumerical

Passage / Comprehension

Two moles of an ideal gas $(γ = 5/3)$ are initially at a temperature of 27°C and occupy a volume of 20 litre. The gas is first expanded at constant pressure until the volume is doubled. It then undergoes reversible adiabatic change until the temperature returns to its initial value. ( $R = 0.082 L - atm / K - mol$ )

The final pressure (in atm) of gas is.

Answer:

0.43

Solution:

We use Boyle's Law, $P_{1} V_{1} = P_{2} V_{2}$ , assuming constant temperature.

Given $P_{1} = 1.2$ atm, $V_{1} = 3.0$ L, $V_{2} = 8.3$ L.

$P_{2} = \frac{P _{1} V _{1}}{V _{2}} = \frac{( 1.2 a t m ) ( 3.0 L )}{8.3 L} \approx 0.43 a t m$

Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:Adiabatic Process/Polytropic

⏱ 2mℹ️ Source: QB

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