Question

Figure 12.8 shows plot of \(\frac{PV}{T}\) 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: T₁>T₂ or T₁<T_{2 ?} 

(c) What is the value of  \(\frac{PV}{T}\) 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 \(\frac{PV}{T}\) at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of  \(\frac{PV}{T}\) (for low pressure high temperature region of the plot) ? (Molecular mass of H₂ = 2.02 u, of O₂ = 32.0 u, R = 8.31 J mo1^–1 K^–1.)

Answer 1

The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the ratio  \(\frac{PV}{T}\) is equal. μR (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.

The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature T₁ is closer to the dotted plot than the curve of the gas at temperature T₂. A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore, T₁ > T₂ is true for the given plot.

The value of the ratio \(\frac{PV}{T}\), where the two curves meet, is μR. This is because the ideal gas equation is given as:

PV = μRT

\(\frac{PV}{T}=μR\)

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 = 1g

R = 8.314 J mole^–1 K^–1

∴ \(\frac{PV}{T}=\frac{1}{32}× 8.314\)

= 0.26 J K^–1

Therefore, the value of the ratio  \(\frac{PV}{T}\)  where the curves meet on the y-axis, is 0.26 J K^–1.

If we obtain similar plots for 1.00 × 10^-3 kg of hydrogen, then we will not get the same at the value of \(\frac{PV}{T}\) 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:

\(\frac{PV}{T}=0.26\,J\,K^{-1}\)

R = 8.314 J mole^–1 K^–1

Molecular mass (M) of H₂ = 2.02 u

\(\frac{PV}{T}=μR \,\,\text{at constant temperature}\)

\(\text{Where, μ=}\frac{m}{M}\)

m = Mass of H₂

∴ \(\frac{PV}{T}×\frac{M}{R}\)

\(=\frac{0.26×2.02}{8.31}\)

= 6.3 × 10^–2 g = 6.3 × 10^–5 kg

Hence, 6.3 × 10^–5 kg of H₂ will yield the same value of  \(\frac{PV}{T}\).