A sample of `2 kg` of helium (assumed ideal) is taken through the process `ABC` and another sample of `2 kg` of the same gas is taken through the proc
A sample of `2 kg` of helium (assumed ideal) is taken through the process `ABC` and another sample of `2 kg` of the same gas is taken through the process `ADC`. Then the temperature of the states `A` and `B` are (Given `R=8.3 "joules//mol"K`):
A. `T_(A)=120.5K,T_(B)=120.5K`
B. `T_(A)=241K,T_(B)=241K`
C. `T_(A)=120.5K, T_(B)=241K`
D.
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Correct Answer - C
`(0.5xx10^(5))/(1.01xx10^(5))xx10xx10^(3)=(2xx1000)/(2)xx0.082T_(A)`
`(1)/(1.01)=0.082T_(A)`
`T_(A)=120.5K`
at constant volume
`(P_(A))/(T_(A))=(P_(6))/(T_(B))`
`T_(B)=(P_(B).T_(A))/(P_(A))=(10)/(5)xx120.5=241K`
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