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Question

$$SO_2$$ and $$NH_3$$ have critical temperatures of $$430.7K$$ and $$405.6K$$, respectively. $$P_C$$ for $$SO_2=77.76$$ atm and $$P_C$$ for $$NH_3=111.5$$ atm. The ratio $$b_{SO_2}:b_{NH_3}$$ is:


A
2:3
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B
1:1
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C
2:1
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D
3:2
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Solution

The correct option is D $$3:2$$
$$T_C=\dfrac{8a}{27bR}$$; $$P_C=\dfrac{a}{27b^2}$$
$$\therefore \dfrac{T_C}{P_C}=\dfrac{8b}{R}$$
$$\Rightarrow b=\dfrac{RT_C}{8P_C}$$
$$\therefore \dfrac{b_{SO_2}}{b_{NH_3}}=\dfrac{430.7\times 111.5}{77.76\times 405.6}=\dfrac{3}{2}$$
$$3:2$$.

Chemistry

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