Question

An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of cross-sectional area 10 cm2 and weight 1 kg (figure). The vessel itself is kept in a big chamber containing air at atmospheric pressure 100 kPa. The length of the gas column is 20 cm. If the chamber is now completely evacuated by an exhaust pump, what will be the length of the gas column? Assume the temperature to remain constant throughout the process.
Figure

Answer 1

The length of the gas column is 2.24 m

Explanation:

Given data:

Cross-sectional area "A" = 100 cm^2 = 10^–3 m

Mass "m" = 1 kg,

Pressure "P" = 100 KPa = 105 Pa

Length of the gas column "ℓ" = 20 cm

Now there are two situations.

• Case I = External pressure exists
• Case II = Internal Pressure does not exist

Solution:

P1V1 = P2V2

(10^5 + 1 x 9.8/10^-3) V = (1 x 9.8/10^-3) V'

(10^5 + 9.8/10^3) A x l = (9.8/10^3) A x l'

l' = 2 x 10^4  + 19.6 x 10^2/9.8/10^3 = 2.24 m

Thus the length of the gas column is 2.24 m

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Answer 2

If the chamber is now completely evacuated by an exhaust pump, the length of the gas column L =2.2 m

Explanation:

Step 1:

Air pressure inside the cylindrical vessel,

$P_{0}=10^{5} \mathrm{Pa}$

$A=10 \mathrm{cm}^{2}=10 \times 10^{-4} \mathrm{m}^{2}$

Pressure due to the weight of the piston $=\frac{m g}{A}=\frac{(1 \times 9.8)}{\left(10 \times 10^{-4}\right)}$

External pressure over the piston after evacuation = 0

$P_{2}=0+9.8 \times 10^{3}$

Step 2:

Now,

$\mathrm{P}_{1} \mathrm{V}_{1}=\mathrm{P}_{2} \mathrm{V}_{2}$

Let L become the column's final length of gas. Then

$\mathrm{V}_{2}=10 \times 10^{-4} \mathrm{L}$

$\begin{array}{l}\left(10^{5}+9.8 \times 10^{3}\right) \times 0.2 \times 10 \times 10^{-4}=9.8 \times 10^{3} \times 10 \times 10^{-4} \mathrm{L} \\\left(10^{5}+9.8 \times 10^{3}\right) \times 2 \times 10^{-4}=98 \times 10^{-1} \mathrm{L}\end{array}$

L =2.2 m

Thus the length of the gas column is 2.2 m