The following graph correctly reflects log P vs log -1-V-for a finite amount of gas at a-constant temperature - True or False- True-False-

The correct option is A True According to Boyle's Law, v ∝1/p ⇒ P=KV Take log on both sides. Log P = log1/V + log K Now, comparing it to Y= mn + c ...

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Standard XII

Chemistry

Boyle's Law

The following...

Question

The following graph correctly reflects log P vs log $(\frac{1}{V})$ for a finite amount of gas at a constant temperature - True or False?

True

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False

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Solution

The correct option is A
True

According to Boyle's Law,

$v \propto \frac{1}{p} \Rightarrow P = K V$

Take log on both sides.

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

Now, comparing it to

Y= mn + c [Straight line]

Therefore, log p and log $\frac{1}{V}$ should be a straight line with slope = k

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Similar questions

The following graph correctly reflects log P vs log $(\frac{1}{V})$ for a finite amount of gas at a constant temperature - True or False?

The following graph correctly reflects log P vs log $(\frac{1}{V})$ for a finite amount of gas at a constant temperature.

K is a constant.

Q. Graph between log p and log1÷V for a fixed amount of a gas at constant temperature

Q. The graph of P vs

\frac{1}{V}

for a finite amount of gas at a constant temperature is a .

Q. The slope of the graph between

log P

and

log V

at constant temperature for a given mass of a gas is:

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The following graph correctly reflects log P vs log (1V) for a finite amount of gas at a constant temperature - True or False?

The correct option is A True According to Boyle's Law, v∝1p ⇒P=KV Take log on both sides. Log P = log1V + log K Now, comparing it to Y= mn + c [Straight line] Therefore, log p and log 1V should be a straight line with slope = k

The following graph correctly reflects log P vs log $(\frac{1}{V})$ for a finite amount of gas at a constant temperature - True or False?

The correct option is A
True

According to Boyle's Law,

$v \propto \frac{1}{p} \Rightarrow P = K V$

Take log on both sides.

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

Now, comparing it to

Y= mn + c [Straight line]

Therefore, log p and log $\frac{1}{V}$ should be a straight line with slope = k