Answer:
The developed pressure is 205.75 bar.
Explanation:
Given that,
Temperature of water =25°C = 273+25 = 298 K
Pressure = 1 bar
Increases temperature = 50°C = 273+50 = 323 K
Value of [tex]\beta = 36.2\times10^{-5}\ K^{-1}[/tex]
Value of [tex]k=4.42\times10^{-5}\ bar^{-1}[/tex]
Specific volume of liquid = 1.0030 cm³/g
We need to calculate the final pressure
Using formula for constant volume change
[tex]\beta(T_{2}-T)_{1})-k(P_{2}-P_{1})=0[/tex]
Put the value into the formula
[tex]36.2\times10^{-5}(323-298)-4.42\times10^{-5}(P_{2}-1)=0[/tex]
[tex]P_{2}=\dfrac{36.2\times10^{-5}(323-298)}{4.42\times10^{-5}}+1[/tex]
[tex]P_{2}=205.75\ bar[/tex]
Hence, The developed pressure is 205.75 bar.
