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. The length of the gas column in the vessel is 20 cm. The atmospheric pressure is 100 kPa. The vessel is now taken into a spaceship revolving round the earth as a satellite. The air pressure in the spaceship is maintained at 100 kPa. Find the length of the gas column in the cylinder.

Answer 1

The length of the gas column in the cylinder is 22cm.

• Given,
• area of a long cylindrical vessel fitted with a frictionless piston = A
• = 10cm^2 = 0.001 m^2
• weight of a long cylindrical vessel fitted with a frictionless
• piston = m
• = 1 Kg
• The atmospheric pressure = P0
• = 100 kPa =10^5 Pa
• The length of the gas column in the cylinder = l'

• Since given, an ideal gas is used,
• we have,
• P1V1 = P2V2
• ⇒

• $(\dfrac{mg}{A}+P0) Al = P0(Al')\\\\(\dfrac{mg}{A}+P0)l = P0(l')\\\\(\dfrac{1 * 9.8}{0.001}+10^5)(0.2) = 10^5(l')\\\\21960 = 10^5(l')\\\\l' = 0.2196$

• ⇒l' = 0.22 m
• ∴l' = 22cm

Answer 2

The length of the gas column in the cylinder is 22 cm.

Explanation:

As we know that ,

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

$\left(\frac{m g}{A}+P_{0}\right) A \ell P_{0} A l$

$\left(\frac{1 \times 9.8}{10 \times 10^{-4}}+10^{5}\right) 0.2=10^{5} \ell^{\prime}$

$\left(9.8 \times 10^{3}+10^{5}\right) \times 0.2=10^{5} \ell^{\prime}$

$\left(9.8 \times 10^{3}+10^{5}\right) \times 0.2=10^{5} \ell^{\prime}$

$109.8 \times 10^{3} \times 0.2=10^{5} l^{'}$

$\ell^{\prime}=\frac{109.8 \times 0.2}{10^{2}}$

   $=0.2196$

   $\approx 22 \mathrm{cm}$

Thus the length of the gas column in the cylinder is 22 cm