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Class 10 Class 12

Kinetic Theory

Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.

The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as,

PV = nRT

where,

R is the universal gas constant = 8.314 J mol^–1 K^–1
n = Number of moles = 1

T = Standard temperature = 273 K

P = Standard pressure = 1 atm = 1.013 × 10⁵ Nm^–2
∴ V =

= 1 × 8.314 × 273 / (1.013 × 10⁵)

= 0.0224 m³
= 22.4 litres

Hence, the molar volume of a gas at STP is 22.4 litres.

219 Views

Molar volume is the volume occupied by molecules of any (ideal) gas at N.T.P. Show that it is 22.4 liters.

Ideal gas equation is given by,
PV = nRT

For 1 mole of gas, n=1

So, PV = RT

At N.T.P, T = 273 K and P = 1.013

\times 10^{5}

Pa
Therefore,
Molar volume is given by,

V =

\frac{RT}{P}

\frac{8.3 \times 273}{1.01 \times 10^{5}} = 22.4 \times 10^{- 3} m^{3} = 22.4 l

398 Views

Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3Å.

Given,

Diameter of an oxygen molecule, d = 3Å

Radius, r =

Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm³
Molecular volume of oxygen gas, V =

πr³N
where,

N is Avogadro’s number = 6.023 x 10²³ molecules/mole

∴ V =

× 3.14 × (1.5 × 10⁸)³ × 6.023 × 10²³

= 8.51 cm³
Ratio of the molecular volume to the actual volume of oxygen =

= 3.8 × 10^-4.

194 Views

Figure 13.8 shows plot of PV/T versus P for 1.00×10^–3kg 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 PV/T where the curves meet on they-axis?

(d) If we obtained similar plots for 1.00 ×10^–3 kg of hydrogen, would we get the same value of

PV over straight T

at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of

PV over straight T

(for low-pressure high-temperature region of the plot)?
(Molecular mass of H₂= 2.02μ, of O₂ = 32.0μ, R = 8.31 J mo1^–1 K^–1.)

(a) The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the ratio

is equal.

μR

where,

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 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.

(c) The value of the ratio

, where the two curves meet, is μR. This is because the ideal gas equation is given as,

PV = μRT

= 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
∴

× 8.314

= 0.26 J K^-1
Therefore, the value of the ratio

, 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

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,

= 0.26 J K^-1
R = 8.314 J mole^–1 K^–1
Molecular mass (M) of H₂ = 2.02 u

= μR, at constant temperature
where, μ = m/M

m = Mass of H₂
∴ m =

= 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

255 Views

An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol^–1 K^–1, molecular mass of O₂ = 32μ).

Volume of oxygen, V₁ = 30 litres = 30 × 10^–3 m³
Gauge pressure, P₁ = 15 atm = 15 × 1.013 × 10⁵ Pa

Temperature, T₁ = 27°C = 300 K

Universal gas constant, R = 8.314 J mole^–1 K^–1
Let the initial number of moles of oxygen gas in the cylinder be n₁.

The gas equation is given as,

P₁V₁ = n₁RT₁
∴ n₁ =

= 18.276

But n₁ = m₁ / M

where,

m₁ = Initial mass of oxygen

M = Molecular mass of oxygen = 32 g

Therefore,

m₁ = n₁M = 18.276 × 32 = 584.84 g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.

Volume, V₂ = 30 litres = 30 × 10^–3 m³
Gauge pressure, P₂ = 11 atm

= 11 × 1.013 × 10⁵ Pa

Temperature, T₂ = 17°C = 290 K

Let n₂ be the number of moles of oxygen left in the cylinder.

The gas equation is given as,

P₂V₂ = n₂RT₂
∴ n₂ = P₂V₂/ RT₂
=

= 13.86

But,

n₂ = m₂ / M

where,

m₂ is the mass of oxygen remaining in the cylinder

∴ m₂ = n₂M = 13.86 × 32 = 453.1 g

The mass of oxygen taken out of the cylinder is given by the relation,

Initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder

= m₁ – m₂
= 584.84 g – 453.1 g

= 131.74 g

= 0.131 kg

Therefore, 0.131 kg of oxygen is taken out of the cylinder.

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Kinetic Theory

Molar volume is the volume occupied by molecules of any (ideal) gas at N.T.P. Show that it is 22.4 liters.