__Question 13.3__:*PV*/

*T*versus P for 1.00×10

^{–3}kg of oxygen gas at two different temperatures.

(b) Which is true:

*T*

_{1 }>

*T*

_{2}or

*T*

_{1}<

*T*

_{2}?

(c) What is the value of

*PV*/

*T*where the curves meet on the

*y-*axis?

(d) If we obtained similar plots for 1.00 ×10

^{–3}kg of hydrogen, would we get the same value of

*PV*/

*T*at the point where the curves meet on the

*y*-axis? If not, what mass of hydrogen yields the same value of

*PV*/

*T*(for low pressure high temperature region of the plot)? (Molecular mass of H

_{2 }= 2.02μ, of O

_{2}= 32.0μ,

*R*= 8.31 J mo1

^{–1}K

^{–1}.)

__Solution__:**(a)**The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the ratio PV/T is equal.

*μ*

*R*(

*μ*is the number of moles and R is the universal gas constant) is a constant quality. It is not dependent on the pressure of the gas.

**(b)**The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature

*T*

_{1}is closer to the dotted plot than the curve of the gas at temperature

*T*

_{2}. A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore,

*T*

_{1}>

*T*

_{2}is true for the given plot.

**(c)**The value of the ratio

*PV*/

*T*, where the two curves meet, is

*μ*

*R*. This is because the ideal gas equation is given as:

*PV*=

*μ*

*RT*

*PV/T =*μR

Where,

*P*is the pressure

*T*is the temperature

*V*is the volume

μ is the number of moles

*R*is the universal constant

Molecular mass of oxygen = 32.0 g

Mass of oxygen = 1 × 10

^{–3}kg = 1 g

*R*= 8.314 J mole

^{–1}K

^{–1}

∴ PV/T = (1/32) X 8.314

= 0.26 J K

^{-1}

Therefore, the value of the ratio

*PV*/

*T*, where the curves meet on the

*y*-axis, is

0.26 J K

^{–1}.

**(d)**If we obtain similar plots for 1.00 × 10

^{–3}kg of hydrogen, then we will not get the same value of

*PV*/

*T*at the point where the curves meet the

*y*-axis. This is because the molecular mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u).

We have:

PV/T = 0.26 J K

^{-1}

*R*= 8.314 J mole

^{–1}K

^{–1}

Molecular mass (

*M*) of H

_{2}= 2.02 u

PV/T = μR at constant temperature

Where, μ = m/M

*m*= Mass of H

_{2}

∴ m = (PV/T) X (M/R)

= 0.26 X 2.02 / 8.31

= 6.3 × 10

^{–2}g = 6.3 × 10

^{–5}kg

Hence, 6.3 × 10

^{–5}kg of H

_{2}will yield the same value of

*PV*/

*T*.

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