Question

Helium gas contained in a piston cylinder assembly undergoes an isentropic polytropic process from the given initial state with P1 = 0.02 bar, T1 = 200 K to the final state with P2 = 0.14 bar. Determine the work and heat transfer (per unit mass) involved in this process.

Answer 1

Heat transfer during the process = 0

Work done during the process = - 371.87 KJ

Step-by-step explanation:

Initial pressure
`P_(1)` = 0.02 bar

Initial temperature
`T_(1)` = 200 K

Final pressure
`P_(2)` = 0.14 bar

Gas constant for helium R = 2.077
`(KJ)/(kg k)`

This is an isentropic polytropic process so temperature - pressure relationship is given by the following formula,

`(T_(2) )/(T_(1) )` =
`[(P_(2) )/(P_(1) ) ]^{(\gamma - 1)/(\gamma) }`

Put all the values in above formula we get,

⇒
`(T_(2) )/(200)` =
`[(0.14 )/(0.02 ) ]^{(1.4 - 1)/(1.4) }`

⇒
`(T_(2) )/(200)` = 1.74

⇒
`T_(2)` = 348.72 K

This is the final temperature of helium.

For isentropic polytropic process heat transfer to the system is zero.

⇒ ΔQ = 0

Work done W = m × (
`T_(1)` -
`T_(2)` ) ×
`(R)/(\gamma - 1)`

⇒ W = 1 × ( 200 - 348.72 ) ×
`(2.077)/(1.4 - 1)`

⇒ W = 371.87 KJ

This is the work done in this process. here negative sign shows that work is done on the gas in the compression of gas.