1 mole of `CO_(2)` gas at 300 K is expanded under adiabatic conditions such that its volume becomes 27 times . What is work done ? (`gamma = 1.33 and

Question

1 mole of `CO_(2)` gas at 300 K is expanded under adiabatic conditions such that its volume becomes 27 times . What is work done ? (`gamma = 1.33 and

1 mole of `CO_(2)` gas at 300 K is expanded under adiabatic conditions such that its volume becomes 27 times . What is work done ? (`gamma = 1.33 and C_(v) = 6 cal mol^(-1) for CO_(2)`)
A. 900 cal
B. 1000 cal
C. 1200 cal
D. 1400 cal

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - c
for adiabatic condition `(T_(2))/(T_(1))=((V_(1))/(v_(2)))^(gamma-1)=(1/27)^(1//3)=1/3`
`T_(2)=300xx1/3 =100K`
thus `T_(2) lt T_(1)` hence cooling takes place due to expansion under adiabatic condition.
`W=DeltaE=-C_(v)(T_(2)-T_(1))`
`[DeltaE=(-)` ve , expansion]
`-6(100-300)=1200 cal`

4 views Answered 2023-02-05 15:29