Question

A vessel of 25 liter capacity contains 10 moles of steam under pressure of 50.3 atm Code
temperature of steam using Vander waal's equation (If for water a = 5 A6 bar 1 mol and
b = 0.031 L mol-!)​

Answer 1

We have, Van der Waals Equation,

$\longrightarrow\sf{\left(P+\dfrac{an^2}{V^2}\right)\big(V-nb\big)=nRT}$

Thus temperature is given by,

$\longrightarrow\sf{T=\dfrac{\left(P+\dfrac{an^2}{V^2}\right)\big(V-nb\big)}{nR}}$

Here,

• $\sf{V=25\ L}$

• $\sf{n=10\ mol}$

• $\sf{P=50.3\ atm=51\ bar}$

• $\sf{a=5.46\ bar\,L^2\,mol^{-2}}$

• $\sf{b=0.031\ L\,mol^{-1}}$

We know the value of $\sf{R}$ is,

$\longrightarrow\sf{R=8.314\ J\,K^{-1}\,mol^{-1}}$

Or,

$\longrightarrow\sf{R=8.314\ m^3\,Pa\,K^{-1}\,mol^{-1}}$

Since $\sf{1\ m^3=10^3\ L,}$

$\longrightarrow\sf{R=8.314\times10^3\ L\,Pa\,K^{-1}\,mol^{-1}}$

And since $\sf{1\ Pa=10^{-5}\ bar,}$

$\longrightarrow\sf{R=8.314\times10^{-2}\ L\,bar\,K^{-1}\,mol^{-1}}$

Hence,

$\longrightarrow\sf{T=\dfrac{\left(P+\dfrac{an^2}{V^2}\right)\big(V-nb\big)}{nR}}$

$\longrightarrow\sf{T=\dfrac{\left(51+\dfrac{5.46\times10^2}{25^2}\right)\big(25-10\times 0.031\big)}{10\times8.314\times10^{-2}}}$

$\longrightarrow\sf{\underline{\underline{T=1540.5\ K}}}$

Hence the temperature is $\bf{1540.5\ K.}$