The graph drawn between pressure and temperature at constant volume for a given mass of different molecular weights $M_{1}$ and $M_{2}$ are the straight lines as shown in the figure then

Question

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Answer 1

From ideal gas equation, PV = nRT = $\frac{w}{M} R T$
Slope of P-T curve = $\frac{w R}{M V}$
Since V = constant and w = constant, we get that the slope is inversely proportional to molecular weight
Steeper the line, more is slope, less is molecular weight.
So, $M_{1} > M_{2}$

Answer 2

From ideal gas equation, PV = nRT =

\frac{w}{M} R T

Slope of P-T curve =

\frac{w R}{M V}

Since V = constant and w = constant, we get that the slope is inversely proportional to molecular weight
Steeper the line, more is slope, less is molecular weight.
So,

M_{1} > M_{2}

The graph drawn between pressure and temperature at constant volume for a given mass of different molecular weights $M_{1}$ and $M_{2}$ are the straight lines as shown in the figure then

$M_{2} > M_{1}$

$M_{1} > M_{2}$

$M_{1} = M_{2}$

$M_{1}^{3} = M_{2}$

From ideal gas equation, PV = nRT = $\frac{w}{M} R T$
Slope of P-T curve = $\frac{w R}{M V}$
Since V = constant and w = constant, we get that the slope is inversely proportional to molecular weight
Steeper the line, more is slope, less is molecular weight.
So, $M_{1} > M_{2}$

The graph drawn between pressure and volume in Boyle's law experiment is shown in figure for two different gases. If the same amount of both the gases has been taken, then the relationship between their molecular weights will be

In Boyles law experiment the graph drawn between pressure and density for a given temperature of different gases are the straight lines then the molecular weights have the relation

The graph drawn between pressure and volume in Boyle's law experiment is shown in figure for different masses of same gas at same temperature then

In Boyles experiment for a given gas at different temperatures the graph drawn between pressure and density are straight lines as shown then:

The P-T graph for the given mass of an ideal gas is shown in figure. Then the volume

Revise with Concepts

Ideal Gas Equation and Absolute Temperature

The graph drawn between pressure and temperature at constant volume for a given mass of different molecular weights M1​ and M2​ are the straight lines as shown in the figure then

M2​>M1​

M1​>M2​

M1​=M2​

M13​=M2​

From ideal gas equation, PV = nRT = Mw​RT Slope of P-T curve = MVwR​ Since V = constant and w = constant, we get that the slope is inversely proportional to molecular weight Steeper the line, more is slope, less is molecular weight. So, M1​>M2​

The graph drawn between pressure and volume in Boyle's law experiment is shown in figure for two different gases. If the same amount of both the gases has been taken, then the relationship between their molecular weights will be

In Boyles law experiment the graph drawn between pressure and density for a given temperature of different gases are the straight lines then the molecular weights have the relation

The graph drawn between pressure and volume in Boyle's law experiment is shown in figure for different masses of same gas at same temperature then

In Boyles experiment for a given gas at different temperatures the graph drawn between pressure and density are straight lines as shown then:

The P-T graph for the given mass of an ideal gas is shown in figure. Then the volume

Revise with Concepts

Ideal Gas Equation and Absolute Temperature

The graph drawn between pressure and temperature at constant volume for a given mass of different molecular weights $M_{1}$ and $M_{2}$ are the straight lines as shown in the figure then

$M_{2} > M_{1}$

$M_{1} > M_{2}$

$M_{1} = M_{2}$

$M_{1}^{3} = M_{2}$

From ideal gas equation, PV = nRT = $\frac{w}{M} R T$
Slope of P-T curve = $\frac{w R}{M V}$
Since V = constant and w = constant, we get that the slope is inversely proportional to molecular weight
Steeper the line, more is slope, less is molecular weight.
So, $M_{1} > M_{2}$