Van der Waals' constants of two gases $X$ and $Y$ are as given:

${a(litre- atm mol}^{- 2})$ ${b(litre mol}^{- 1})$
Gas $X$ $5.6$ $0.065$
Gas $Y$ $5.1$ $0.012$

What is correct about the two gases?

T_{c} (X) > T_{c} (Y)

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T_{c} (X) = T_{c} (Y)

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V_{c} (X) > V_{c} (Y)

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V_{c} (Y) > V_{c} (X)

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Solution

The correct option is B $V_{c} (X) > V_{c} (Y)$
$V_{c} (X) = 3 \times b_{X} = 3 \times 0.065 = 0.195 litre/mole$

$V_{c} (Y) = 3 \times b_{Y} = 3 \times 0.012 = 0.036 litre/mole$

$T_{c} = \frac{8 a}{27 R b}$

The ratio of a and b of X $= 86.15$

The ratio of a and b of Y $= 425$

Therefore, $T_{c} (X) < T_{c} (Y)$ and $V_{c} (X) > V_{c} (Y)$

Hence, the correct option is $C$

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	${a(litre- atm mol}^{- 2})$	${b(litre mol}^{- 1})$
Gas $X$	$5.6$	$0.065$
Gas $Y$	$5.1$	$0.012$

Van der Waals' constants of two gases X and Y are as given:a(litre- atm mol−2)b(litre mol−1)Gas X5.60.065Gas Y5.10.012What is correct about the two gases?

The correct option is B Vc(X)>Vc(Y)Vc(X)=3×bX=3×0.065=0.195 litre/moleVc(Y)=3×bY=3×0.012=0.036 litre/moleTc=8a27RbThe ratio of a and b of X=86.15The ratio of a and b of Y=425Therefore, Tc(X)<Tc(Y) and Vc(X)>Vc(Y)Hence, the correct option is C

Van der Waals' constants of two gases $X$ and $Y$ are as given:

${a(litre- atm mol}^{- 2})$ ${b(litre mol}^{- 1})$
Gas $X$ $5.6$ $0.065$
Gas $Y$ $5.1$ $0.012$

What is correct about the two gases?