Question 965: Thermodynamics & KTG

Question

An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressures at temperature $27^{\circ} \mathrm{C}$. The mass of the oxygen withdrawn from the cylinder is nearly equal to:
[Given, $R=\frac{100}{12} \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, and molecular mass of $\mathrm{O}_2=32,1$ atm pressure $=1.01 	imes 10^5 \mathrm{~N} / \mathrm{m}$]

Accepted Answer

To find the mass of oxygen withdrawn from the cylinder, we start by calculating the number of moles left in the cylinder after some oxygen is withdrawn. We use the ideal gas law in the form:
$ n = \frac{PV}{RT} $
Substituting the given values:
$ P = 11 $ atm converts to $ 11 	imes 1.01 	imes 10^5 \, 	ext{N/m}^2 $
$ V = 30 \, 	ext{liters} = 30 	imes 10^{-3} \, 	ext{m}^3 $
$ R = \frac{100}{12} \, 	ext{J/mol K} $
$ T = 27^\circ 	ext{C} = 300 \, 	ext{K} $
We calculate the moles after oxygen has been withdrawn:
$ n = \frac{12 	imes 1.01 	imes 10^5 \, 	ext{N/m}^2 	imes 30 	imes 10^{-3} \, 	ext{m}^3}{\left(\frac{100}{12}ight) 	imes 300} $
Simplifying the expression:
$ n = \frac{12 	imes 1.01 	imes 12}{10} = 14.54 \, 	ext{moles} $
Next, determine the moles of oxygen removed:
$ 	ext{Moles removed} = 18.20 - 14.54 = 3.656 \, 	ext{moles} $
Finally, convert the moles removed into mass:
$ 	ext{Mass removed} = 3.656 	imes 32 = 116.99 \, 	ext{g} = 0.116 \, 	ext{kg} $
Thus, the mass of the oxygen withdrawn from the cylinder is approximately 0.116 kg.

Answer

0.116 kg

Answer

0.156 kg

Answer

0.125 kg

Answer

0.144 kg

Thermodynamics & KTG

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