Question

A glass tube of capacity 3 litre is heated at sea level from 0°C to 114°C with its mouth open. Find the mass of air that expelled. Assuming density of air at NTP is 1.29 x 10 Kg/m³​

Answer 1

The mass of air that expelled is:

1. A glass tube of capacity 3 litre is heated at sea level from 0°C to 114°C with its mouth open.
2. Let us assume volume of air expelled be V and final volume be V1 + V.
3. We are given, T1 = 273K, T2 = 387K and V1 = 3L
4. Density has to be 1.29 Kg/m³
5. At constant pressure, according to charles law, volume of a gas is directly proportional to Temperature.

.•. V2/V1 = T2/T1

V1 + V/V1 = T2/T1

(V1 + V)×T1 = V1×T2

V = V1×T2 - V1×T1/T1

= (3×387 - 3×273)/273

= 3×114/273

= 114/91

= 1.25 L = 1.25×10-3 m^-3

6. For mass of air expelled, we know that

density = mass/volume

mass = density×volume

= 1.29×1.25×10^-3 Kg

= 1.61×10^-3

Answer 2

The mass of air that expelled when heated a glass tube is $1.61 \times 10^{-3} \text{kg}$

Explanation:

Given data

Capacity of the glass tube $(\text{V}_1)$ = 3 Litre

Initial temperature = 0° C = 273 K

Final temperature = 114° C = 387 K

Density = 1.29 Kg/m³

Find the mass of air expelled when the glass tube is heated.

Let us consider,

Expelled volume of air = V

$\text{Final volume} \ \text{V}_2 = \text{V}_1 + \text{V}$

Charles law - The volume of a gas is directly proportional to temperature, when pressure is constant.

                                 $\text{V} \propto \text{T}$

$\Rightarrow \frac{\text{V}_2}{\text{V}_1} = \frac{\text{T}_2}{\text{T}_1}$

$\Rightarrow \frac{\text{V}_1 + \text{V}}{\text{V}_1} = \frac{\text{T}_2}{\text{T}_1}$

$\Rightarrow (\text{V}_1+ \text{V}) \times \text{T}_1 = \text{T}_2 \times {\text{V}_1}$

$\Rightarrow \text{V} = \frac{ \text{V}_1 {\text{T}_2 - \text{V}_1\text{T}_1}}{\text{T}_1}$

Substitute the respective values in above equation.

$\text{V} = \frac{ (3 \times 387)- ( 3 \times 273 )}{273}$

$\text{V} = \frac{ 114 }{91}$

V = 1.25 L

$\text{V} = 1.25 \times 10^{-3} m^{3}$

$\text{Density} = \frac{\text{mass}}{\text{volume}}$

Then,

Mass = Density × Volume

$\text{Mass} = 1.29 \times 1.25 \times 10^{-3}$

$\text{Mass} = 1.61 \times 10^{-3} \text{kg}$

Therefore the mass of the air expelled is $1.61 \times 10^{-3} \text{kg}$

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