Respuesta :
Answer: The volume of the vessel is [tex]0.1542m^3[/tex] and total internal energy is 162.0 kJ.
Explanation:
- To calculate the volume of water, we use the equation given by ideal gas, which is:
[tex]PV=nRT[/tex]
or,
[tex]PV=\frac{m}{M}RT[/tex]
where,
P = pressure of container = 700 kPa
V = volume of container = ? L
m = Given mass of R-134a = 3.98 kg = 3980 g (Conversion factor: 1kg = 1000 g)
M = Molar mass of R-134a = 102.03 g/mol
R = Gas constant = [tex]8.31\text{L kPa }mol^{-1}K^{-1}[/tex]
T = temperature of container = [tex]60^oC=[60+273]K=333K[/tex]
Putting values in above equation, we get:
[tex]700kPa\times V=\frac{3980g}{102.03g/mol}\times 8.31\text{L kPa }\times 333K\\\\V=154.21L[/tex]
Converting this value into [tex]m^3[/tex], we use the conversion factor:
[tex]1m^3=1000L[/tex]
So, [tex]\Rightarrow (\frac{1m^3}{1000L})\times 154.21L[/tex]
[tex]\Rightarrow 0.1542m^3[/tex]
- To calculate the internal energy, we use the equation:
[tex]U=\frac{3}{2}nRT[/tex]
or,
[tex]U=\frac{3}{2}\frac{m}{M}RT[/tex]
where,
U = total internal energy
m = given mass of R-134a = 3.98 kg = 3980 g (Conversion factor: 1kg = 1000g)
M = molar mass of R-134a = 102.03 g/mol
R = Gas constant = [tex]8.314J/K.mol[/tex]
T = temperature = [tex]60^oC=[60+273]K=333K[/tex]
Putting values in above equation, we get:
[tex]U=\frac{3}{2}\times \frac{3980g}{102.03g/mol}\times 8.314J/K.mol\times 333K\\\\U=161994.6J[/tex]
Converting this into kilo joules, we use the conversion factor:
1 kJ = 1000 J
So, 161994.6 J = 162.0 kJ
Hence, the volume of the vessel is [tex]0.1542m^3[/tex] and total internal energy is 162.0 kJ.
