At relatively high pressure,the van der Waals equation becomes?
- A$PV_m = RT - Pb$
- B$PV_m = RT - \frac{a}{V_m}$
- C$PV_m = RT + Pb$
- D$PV_m = RT - \frac{a}{V_m^2}$
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Similar Questions
Consider the following table:
Gas $a / (kPa \cdot dm^6 \cdot mol^{-2})$ $b / (dm^3 \cdot mol^{-1})$ $A$ $642.32$ $0.05196$ $B$ $155.21$ $0.04136$ $C$ $431.91$ $0.05196$ $D$ $155.21$ $0.4382$
$a$ and $b$ are van der Waals constants. The correct statement about the gases is:
| Gas | $a / (kPa \cdot dm^6 \cdot mol^{-2})$ | $b / (dm^3 \cdot mol^{-1})$ |
|---|---|---|
| $A$ | $642.32$ | $0.05196$ |
| $B$ | $155.21$ | $0.04136$ |
| $C$ | $431.91$ | $0.05196$ |
| $D$ | $155.21$ | $0.4382$ |
$a$ and $b$ are van der Waals constants. The correct statement about the gases is:
DifficultJEE MAIN 2019
View SolutionUnder which conditions does the $van \ der \ Waals$ equation reduce to the ideal gas equation?
Easy
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The correct increasing order of the value of van der Waals's constant $a$ for $CH_4, O_2$ and $H_2$ gases is
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