### Q.3:- Figure 13.8 shows plot of *PV*/*T *versus P for 1.00×10^{–3} kg of oxygen gas at two different temperatures.

### (a) What does the dotted plot signify?

(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}.)

**Answer:-**

(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) × 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* = μ*R* at constant temperature

Where, μ = *m*/*M*

*m* = Mass of H_{2}

∴ *m* = (*PV*/*T*) × (*M*/*R*)

= 0.26 × 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*.