Answer:
Amount of air left in the cylinder=m[tex]_{2}[/tex]=0.357 Kg
The amount of heat transfer=Q=0
Explanation:
Given
Initial pressure=P1=300 KPa
Initial volume=V1=0.2[tex]m^{3}[/tex]
Initial temperature=T[tex]_{1}[/tex]=20 C
Final Volume=[tex]V_{2}[/tex]=0.1 [tex]m^{3}[/tex]
Using gas equation
[tex]m_{1}=((P_{1}*V_{1})/(R*T_{1}))[/tex]
m1==(300*0.2)/(.287*293)
m1=0.714 Kg
Similarly
m2=(P2*V2)/R*T2
m2=(300*0.1)/(0.287*293)
m2=0.357 Kg
Now calculate mass of air left,where me is the mass of air left.
me=m2-m1
me=0.715-0.357
mass of air left=me=0.357 Kg
To find heat transfer we need to apply energy balance equation.
[tex]Q=(m_{e}*h_{e})+(m_{2}*h_{2})-(m_{1}*h_{1})[/tex]
Where me=m1-m2
And as the temperature remains constant,hence the enthalpy also remains constant.
h1=h2=he=h
Q=(me-(m1-m2))*h
me=m1-me
Thus heat transfer=Q=0
