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

Answer:

Heat transfer during the process = 0

Work done during the process = - 371.87 KJ

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 $\frac{KJ}{kg k}$

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

$\frac{T_{2} }{T_{1} }$ = $[\frac{P_{2} }{P_{1} } ]^{\frac{\gamma - 1}{\gamma} }$

Put all the values in above formula we get,

⇒ $\frac{T_{2} }{200}$ = $[\frac{0.14 }{0.02 } ]^{\frac{1.4 - 1}{1.4} }$

⇒  $\frac{T_{2} }{200}$ = 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 × ( $T_{1}$ - $T_{2}$ ) × $\frac{R}{\gamma - 1}$

⇒ W = 1 × ( 200 - 348.72 ) × $\frac{2.077}{1.4 - 1}$

⇒ 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.