Answer:
a) [tex](dP)_{v} = 9.692 kPa[/tex]
b) [tex](dP)_{T} = -9.692 kPa[/tex]
c) dP = 0 Pa
Explanation:
The specifies equation is :
[tex]dz = (\frac{\delta z}{\delta x}) _{y} dx + (\frac{\delta z}{\delta y}) _{x} dy[/tex]
Note that:
[tex]dP = \frac{R}{v} dT - \frac{RT}{v^{2} } dV[/tex]
1% increase in temperature at specific volume:
[tex]dT = \frac{0.01}{1} *350\\dT = 3.5 K[/tex]
a) Change in pressure of helium at constant volume:
[tex](dP)_{v} = \frac{R}{v} dT[/tex]
R = 2.0769 kJ/kg-K
dT = 3.5 K
v = 0.75 m³/kg
[tex](dP)_{v} = \frac{2.0769}{0.75} * 3.5\\(dP)_{v} = 9.692 kPa[/tex]
b)
dv = (1%/100%) *0.75
dv = 0.0075 m³/kg
Change in pressure of helium at constant temperature:
[tex](dP)_{T} = \frac{-RT}{v^{2} } dv[/tex]
R = 2.0769 kJ/kg-K
T = 350 K
v = 0.75 m³/kg
dv = 0.0075 m³/kg
[tex](dP)_{T} = \frac{-(2.0769*350)}{0.75^{2} } *0.0075\\(dP)_{T} = -9.692 kPa[/tex]
c) The change in pressure of helium :
[tex]dP = (dP)_{v} + (dP)_{T}[/tex]
dP = 9.692 - 9.692
dP = 0
