A rigid vessel of diameter $0\mathrm{.}5m$ contains $0\mathrm{.}92kg$ of air at $720K$ and $3$ bars.

Considering the air to behave as an ideal gas $R=287$

a$.$ Determine the specific volume of the gas and the volume of the vessel.

volume gas $=pV=mRTV=\frac{mRT}{p}=\frac{0\mathrm{.}92287720}{3{10}^5}$

$=0\mathrm{.}634kg{m}^{-3}$

volume vessel $=pv=RTv=\frac{RT}{p}=\frac{287720}{3{10}^5}=0\mathrm{.}689m^{3}k{g}^{-1}$

b$.$ Some of the air is vented isothermally to an empty cylinder of diameter

$0\mathrm{.}4m$ which has a $900kg$ mass on the piston as shown in Figure $Q1.$ If

the piston rises to a height of $0\mathrm{.}3m,$ determine the mass of gas in the

cylinder.

rigure $41$

Note: we are assuming here, that the pressure is maintained in the cylinder

due to the mass.$|$

c$.$A valve isolates the cylinder from the rigid vessel and $20kJ$ of heat is

added to the cylinder. Calculate the new volume of the cylinder and the

work done firstly as an isobaric process and isothermal processes.

Taking $C_p=1\mathrm{.}02kJk{g}^{-1}{K}^{-1}$ and $C_C=0\mathrm{.}718kJk{g}^{-1}{K}^{-1}.$

d$.20kJ$ of heat is now added to the rigid tank. What is the new

temperature?