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<T2 ?

(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 $\frac{PV}{T}$.