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The plot of PV vs. P at a particular temperature is called isobar.

(a) True

(b) False

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In order to solve this question, we need to first understand Boyleâ€™s law. According to Boyleâ€™s law, the volume of a given mass of a gas is inversely proportional to its pressure. That is:

$ V\propto \cfrac { 1 }{ P } $

Where V is the volume of the gas and P is the pressure exerted by the gas. Therefore:

$ V = k\times \cfrac { 1 }{ P } $

Where k is the proportionality constant. Hence,

$ PV = k$

This means the product of the pressure and volume for a given amount of a gas is a constant at constant temperature.

Boyleâ€™s law proves that gases are compressible. If we increase the pressure, the gases will become denser. Hence we can say that at a constant temperature, the density of a gas is directly proportional to its pressure for a fixed mass of a gas.

So, if m is the mass of a gas, V is its volume and P is its pressure, then its density d will be:

$ d = \cfrac { m }{ V } $

Now, according to Boyleâ€™s law, $ PV = k$ or $ V = k\times \cfrac { 1 }{ P } $

Putting this value of V in the density formula,

$d = \cfrac { m }{ \cfrac { k }{ P } } $

$\Rightarrow d = \cfrac { m }{ k } \times P = { k }^{ \prime }P$

Where ${ k }^{ \prime } = \cfrac { m }{ k }$, another constant.

Therefore at higher altitudes the air is less dense since at higher altitudes the pressure is very low. This results in altitude sickness.

Now if we look at the ideal gas equation which states that the product of the volume and pressure of a gas is equal to the product of its number of moles, the temperature and the gas constant:

PV = nRT

If we taking a fixed amount of a gas at constant temperature, this will imply that:

PV = k

Where k = nRT, a constant

This is actually Boyle's law.

In thermodynamics, an isobaric process is a process which is carried out at a constant pressure. So, if we take a fixed mass of an ideal at a constant pressure then its volume will be directly proportional to the temperature which is called the Charlesâ€™ law:

$ V\propto T$

$ \Rightarrow V = kT$ where k is the constant of proportionality.

Since in the question we are also taking the temperature to be a constant, hence

$ V = { k }^{ \prime }$

Where ${ k }^{ \prime }$ = kT, hence if we multiply both sides with pressure, we will get:

$ \Rightarrow PV = { k }^{ \prime }P$

Since the process is also isobaric, i.e. the pressure is constant therefore:

$\Rightarrow PV = { k }^{ \prime \prime }$

Where ${ k }^{ \prime \prime } = { k }^{ \prime }P$

This implies that the plot of PV vs. P at a particular temperature for a fixed amount of a gas will be a straight line parallel to the x-axis and it will be an isobaric process.