Answer :

Explanation: We have to consider 2 cases here and find the masses of O2 in each case and then by calculating the difference of those masses gives the the mass of oxygen taken out of the cylinder.


Given:


Initially, Case1


V1 = 30 litres = 0.03 m3


P1 = 15 atm = 15 × 1.01 × 105 Pa


T1 = 27+273 = 300 K


If n1 is the moles O2 in the cylinder, then by using


P1V1 = n1RT1 ---( Ideal Gas Law )


n1 = =


n1 = 18.276 moles


Molecular weight of O2 (M) = 32 g


Initial mass of cylinder (m1) = n1M


m1 = n1M


m1 = (18.276 moles) × ( 32 g)


m1 = 584.84 g


Now,


Let n2 be the moles of O2 left in the cylinder


V2 = 30 litres = 0.03 m3


P2 = 11 atm = 11 × 1.01 × 105 Pa


T2 = 17+273 = 290 K


Thus n2 = =


n1 = 13.86 moles


Therefore, m2 be the final mass of O2 in the cylinder


m2 = n2M


m2 = (13.86 moles) × ( 32 g)


m2 = 453.1 g


the mass of oxygen taken out of the cylinder


Δm = m1-m2


Δm = 584.84 g - 453.1 g


Δm = 131.74 g


The mass of oxygen taken out of the cylinder is 131.74 g.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

A box of 1.00m3 iPhysics - Exemplar

Explain why
</
Physics - Exemplar

Consider a rectanPhysics - Exemplar

Consider an idealPhysics - Exemplar

Two molecules of Physics - Exemplar

When an ideal gasPhysics - Exemplar

In a diatomic molPhysics - Exemplar

An insulated contPhysics - Exemplar

Diatomic moleculePhysics - Exemplar

ABCDEFGH is a holPhysics - Exemplar