One mole of an ideal monatomic gas undergoes the process `P=alphaT^(1//2)`, where `alpha` is constant . If molar heat capacity of the gas is `betaR1`

Question

One mole of an ideal monatomic gas undergoes the process `P=alphaT^(1//2)`, where `alpha` is constant . If molar heat capacity of the gas is `betaR1`

One mole of an ideal monatomic gas undergoes the process `P=alphaT^(1//2)`, where `alpha` is constant . If molar heat capacity of the gas is `betaR1` when `R` = gas constant then find the value of `beta`.

Monjur Alahi

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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 - 2
`PalphaT^(1//2)` and `PV = nRt implies PV^(-1)` = constant `therefore C=C_(v) + (R)/(1-(-1)) = (3R)/(2) + (R)/(2) = 2R`

4 views Answered 2023-02-05 15:29