you want to cool a storage cylinder containing 10 moles of compressed gas from 30oc to 20oc. for which kind of gas would this be easiest?

you want to cool a storage cylinder containing 10 moles of compressed gas from 30oc to 20oc. for which kind of gas would this be easiest?

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you want to cool a storage cylinder containing 10 moles of compressed gas from 30oc to 20oc. for which kind of gas would this be easiest?

The answer to this question is A – Monoatomic Gas Explanation- A monoatomic gas has only one atom and its degree of freedom is three (3). It means it can only have translation motion in any of the three axes. Diatomic gas has 5 degrees of freedom

you want to cool a storage cylinder containing 10 moles of compressed gas from 30oc to 20oc. for which kind of gas would this be easiest?

Volume of oxygen, $V_{1} =$ 30 litres $= 30 \times 1 0^{-3} m^{3}$

Gauge pressure,

P_{1} =

15 atm

= 15 \times 1.013 \times 1 0^{5} P a

Temperature,

T_{1} = 2 7^{o} C = 300 K

Universal gas constant,

R = 8.314 J m o l^{-1} K^{-1}

Let the initial number of moles of oxygen gas in the cylinder be n₁.
The gas equation is given as:

P_{1} V_{1} = n_{1} R T_{1}

∴ n_{1} = P_{1} V_{1} / R T_{1}

= (15.195 \times 1 0^{5} \times 30 \times 1 0^{-3}) / (8.314 \times 300) = 18.276

But

n_{1} = m_{1} / M

Where,

m_{1} =

Initial mass of oxygen
M = Molecular mass of oxygen = 32 g

∴ m_{1} = N_{1} M = 18.276 \times 32 = 584.84 g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.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

V_{2} =

30 litres

= 30 \times 1 0^{-3} m^{3}

Gauge pressure,

P_{2} =

11 atm

= 11 \times 1.013 \times 1 0^{5} P a

Temperature,

T_{2} = 1 7^{o} C = 290 K

Let

n_{2}

be the number of moles of oxygen left in the cylinder.
The gas equation is given as:

P_{2} V_{2} = n_{2} R T_{2}

∴ n_{2} = P_{2} V_{2} / R T_{2}

= (11.143 \times 1 0^{5} \times 30 \times 1 0^{-30}) / (8.314 \times 290) = 13.86

But

n_{2} = m_{2} / M

Where,

m_{2}

is the mass of oxygen remaining in the cylinder

∴ m_{2} = n_{2} \times M = 13.86 \times 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

= m_{1} - m_{2}

= 584.84 g - 453.1 g

= 131.74 g

= 0.131 k g

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

you want to cool a storage cylinder containing 10 moles of compressed gas from 30oc to 20oc. for which kind of gas would this be easiest?

you want to cool a storage cylinder containing 10 moles of compressed gas from 30oc to 20oc. for which kind of gas would this be easiest?

Volume of oxygen, V1​= 30 litres =30×10−3m3

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Volume of oxygen, $V_{1} =$ 30 litres $= 30 \times 1 0^{-3} m^{3}$