Answer: The volume of the container is [tex]2.8497m^3[/tex]
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 = 200 kPa
V = volume of container = ? L
m = Given mass of water = 2.61 kg = 2610 g (Conversion factor: 1kg = 1000 g)
M = Molar mass of water = 18 g/mol
R = Gas constant = [tex]8.31\text{L kPa }mol^{-1}K^{-1}[/tex]
T = temperature of container = [tex]200^oC=[200+273]K=473K[/tex]
Putting values in above equation, we get:
[tex]200kPa\times V=\frac{2610g}{18g/mol}\times 8.31\text{L kPa }\times 473K\\\\V=2849.7L[/tex]
Converting this into cubic meter, we use the conversion factor:
[tex]1m^3=1000L[/tex]
So, [tex]\Rightarrow \frac{1m^3}{1000L}\times 2849.7L[/tex]
[tex]\Rightarrow 2.8497m^3[/tex]
Hence, the volume of the container is [tex]2.8497m^3[/tex]
