Answer:
Heat transfer during the process = 0
Work done during the process = - 371.87 KJ
Explanation:
Initial pressure [tex]P_{1}[/tex] = 0.02 bar
Initial temperature [tex]T_{1}[/tex] = 200 K
Final pressure [tex]P_{2}[/tex] = 0.14 bar
Gas constant for helium R = 2.077 [tex]\frac{KJ}{kg k}[/tex]
This is an isentropic polytropic process so temperature - pressure relationship is given by the following formula,
[tex]\frac{T_{2} }{T_{1} }[/tex] = [tex][\frac{P_{2} }{P_{1} } ]^{\frac{\gamma - 1}{\gamma} }[/tex]
Put all the values in above formula we get,
⇒ [tex]\frac{T_{2} }{200}[/tex] = [tex][\frac{0.14 }{0.02 } ]^{\frac{1.4 - 1}{1.4} }[/tex]
⇒ [tex]\frac{T_{2} }{200}[/tex] = 1.74
⇒ [tex]T_{2}[/tex] = 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 × ( [tex]T_{1}[/tex] - [tex]T_{2}[/tex] ) × [tex]\frac{R}{\gamma - 1}[/tex]
⇒ W = 1 × ( 200 - 348.72 ) × [tex]\frac{2.077}{1.4 - 1}[/tex]
⇒ 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.
