The dimensions of - a- in Vanderwaal-s equation P- a V 2 V-b -RT is- V-volume- P-Pressure- R-Universal gas constant- T- Temperature

The correct option is D M ^ 1 L ^ 5 T ^ 2 ( P+ aV^2) ( v b ) = RT aV^2 #160; should have the #160; some #160; dimensions #160; as ...

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Standard XII

Physics

1st Law of Thermodynamics

The dimension...

Question

The dimensions of ' $a$ ' in Vanderwaal's equation $(P + \frac{a}{V^{2}}) (V - b) = R T$ is:
( $V$ -volume, $P$ -Pressure, $R$ -Universal gas constant, $T$ - Temperature)

M^{1} L^{- 1} T^{- 2}

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M^{1} L^{5} T^{- 2}

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M^{0} L^{3} T^{0}

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M^{0} L^{6} T^{0}

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Solution

The correct option is D $M^{1} L^{5} T^{- 2}$
$(P + \frac{a}{V^{2}}) (v - b) = R T$
$\frac{a}{V^{2}}$ should have the some dimensions as $P$

$\frac{a}{V^{2}} = M^{1} L^{- 1} T^{- 2}$
$a = (M^{1} L^{- 1} T^{- 2}) (L^{6}) = M^{1} L^{5} T^{- 2}$

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Similar questions

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The dimensions of 'a' in Vanderwaal's equation (P+aV2)(V−b)=RT is:(V-volume, P-Pressure, R-Universal gas constant, T- Temperature)

The correct option is D M1L5T−2(P+aV2)(v−b)=RTaV2 should have the some dimensions as PaV2=M1L−1T−2a=(M1L−1T−2)(L6)=M1L5T−2

The dimensions of ' $a$ ' in Vanderwaal's equation $(P + \frac{a}{V^{2}}) (V - b) = R T$ is:
( $V$ -volume, $P$ -Pressure, $R$ -Universal gas constant, $T$ - Temperature)

The correct option is D $M^{1} L^{5} T^{- 2}$
$(P + \frac{a}{V^{2}}) (v - b) = R T$
$\frac{a}{V^{2}}$ should have the some dimensions as $P$

$\frac{a}{V^{2}} = M^{1} L^{- 1} T^{- 2}$
$a = (M^{1} L^{- 1} T^{- 2}) (L^{6}) = M^{1} L^{5} T^{- 2}$