Question

One mole each of gas (1) and gas (2) occupies 0.5 and 2 L, respectively at ${27}^{o}C$. The compressibility factors of the two gases at this temperature are 0.8 and 0.2 respectively. If $b=0.04Lmo{l}^{-1}$ for both the gases, the ratio of van der Waal’s gas constant $a_1:a_2$ for the gases is

• A. 0.4
• B. 0.1
• C. 0.2
• D. 1

Answer 1

The correct option is B 0.1

Given, $n_1=n_2=1,Z_1=0.8,Z_2=0.2,V_{m_1}=0.5L,V_{m_2}=2L,b=0.04L$

For a van der Wall’s gas -

$Z=\frac{Vm}{(Vm-b)}-\frac{a}{V_{m}RT}$

By putting the given values in the above formula we get,

$\frac{a_1}{a_2}=\frac{\left(\right[V_{m_1}/(V_{m_1}-b)]-Z_1)V_{m_1}}{\left(\right[V_{m_2}/(V_{m_2}-b)]-Z_2)V_{m_2}}$

$\frac{a_1}{a_2}=\frac{0.132\times 1.96}{0.46\times 1.608\times 4}=0.08\approx 0.1$

Option B is correct.