The intercept of the line drawn for log P -P in atm- and log1-V-V in the litre- for 1 mole of an ideal gas at- 27- C is equal tolog 2-463-log 22-4-log 24-63-log 2-24-

The correct option is C log 24.63 PV = K or log P = log K + log 1/V Also PV = RT = 0.0821 × 300 = 24.63 ∴ K= 24.63 ...

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Standard XII

Chemistry

Ideal Gas Equation

The intercept...

Question

The intercept of the line drawn for log P (P in atm) and $l o g \frac{1}{V}$ (V in the litre) for 1 mole of an ideal gas at $27^{\circ}$ C is equal to

log 2.463

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log 24.63

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log 22.4

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log 2.24

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Solution

The correct option is B
log 24.63

PV = K

or log P = log K + log $\frac{1}{V}$

Also PV = RT = 0.0821 $\times$ 300

= 24.63

$∴$ K= 24.63

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The intercept of the line drawn for log P (P in atm) and log1V (V in the litre) for 1 mole of an ideal gas at 27∘ C is equal to

The correct option is B log 24.63 PV = K or log P = log K + log 1V Also PV = RT = 0.0821 × 300 = 24.63 ∴ K= 24.63

The intercept of the line drawn for log P (P in atm) and $l o g \frac{1}{V}$ (V in the litre) for 1 mole of an ideal gas at $27^{\circ}$ C is equal to

The correct option is B
log 24.63

PV = K

or log P = log K + log $\frac{1}{V}$

Also PV = RT = 0.0821 $\times$ 300

= 24.63

$∴$ K= 24.63