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ChemistryThermodynamics - IAdiabatic Process/PolytropicMedium2 minQB

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10 moles of $CH_{4}$ gas are adiabatically compressed from 400 K, 2 atm to 600 against constant external pressure 10 atm. Final pressure of $CH_{4}$ gas is

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Answer:

Solution:

For adiabatic process
$Δ U = w$
$n C_{v, m} (T_{2} - T_{1}) = - P_{ext} (\frac{n R T _{2}}{P _{2}} - \frac{n R T _{1}}{P _{1}})$
$n \times 3 R (600 - 400) = - 10 \times n R (\frac{600}{P _{2}} - \frac{400}{2})$
$60 = 200 - \frac{600}{P _{2}}$
$P_{2} = \frac{30}{7} atm$

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

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