<u>Answer:</u> The volume of the vessel is
and total internal energy is 162.0 kJ.
<u>Explanation:</u>
- To calculate the volume of water, we use the equation given by ideal gas, which is:
![PV=nRT](https://tex.z-dn.net/?f=PV%3DnRT)
or,
![PV=\frac{m}{M}RT](https://tex.z-dn.net/?f=PV%3D%5Cfrac%7Bm%7D%7BM%7DRT)
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 = ![8.31\text{L kPa }mol^{-1}K^{-1}](https://tex.z-dn.net/?f=8.31%5Ctext%7BL%20kPa%20%7Dmol%5E%7B-1%7DK%5E%7B-1%7D)
T = temperature of container = ![60^oC=[60+273]K=333K](https://tex.z-dn.net/?f=60%5EoC%3D%5B60%2B273%5DK%3D333K)
Putting values in above equation, we get:
![700kPa\times V=\frac{3980g}{102.03g/mol}\times 8.31\text{L kPa }\times 333K\\\\V=154.21L](https://tex.z-dn.net/?f=700kPa%5Ctimes%20V%3D%5Cfrac%7B3980g%7D%7B102.03g%2Fmol%7D%5Ctimes%208.31%5Ctext%7BL%20kPa%20%7D%5Ctimes%20333K%5C%5C%5C%5CV%3D154.21L)
Converting this value into
, we use the conversion factor:
![1m^3=1000L](https://tex.z-dn.net/?f=1m%5E3%3D1000L)
So, ![\Rightarrow (\frac{1m^3}{1000L})\times 154.21L](https://tex.z-dn.net/?f=%5CRightarrow%20%28%5Cfrac%7B1m%5E3%7D%7B1000L%7D%29%5Ctimes%20154.21L)
![\Rightarrow 0.1542m^3](https://tex.z-dn.net/?f=%5CRightarrow%200.1542m%5E3)
- To calculate the internal energy, we use the equation:
![U=\frac{3}{2}nRT](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B3%7D%7B2%7DnRT)
or,
![U=\frac{3}{2}\frac{m}{M}RT](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B3%7D%7B2%7D%5Cfrac%7Bm%7D%7BM%7DRT)
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 = ![8.314J/K.mol](https://tex.z-dn.net/?f=8.314J%2FK.mol)
T = temperature = ![60^oC=[60+273]K=333K](https://tex.z-dn.net/?f=60%5EoC%3D%5B60%2B273%5DK%3D333K)
Putting values in above equation, we get:
![U=\frac{3}{2}\times \frac{3980g}{102.03g/mol}\times 8.314J/K.mol\times 333K\\\\U=161994.6J](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B3%7D%7B2%7D%5Ctimes%20%5Cfrac%7B3980g%7D%7B102.03g%2Fmol%7D%5Ctimes%208.314J%2FK.mol%5Ctimes%20333K%5C%5C%5C%5CU%3D161994.6J)
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
and total internal energy is 162.0 kJ.