A 1 L flask contains 32 g O2 gas at 27° C . What mass of O2 must be released to reduce the pressure in the flask to 12.315 atm?

Question

A 1 L flask contains 32 g O2 gas at 27° C . What mass of O2 must be released to reduce the pressure in the flask to 12.315 atm? (A) 8 g (B) 16 g (C) 24 g (D) 32 g

Tardigrade Admin · Accepted Answer

pV=nRT=(w/m)Rt w=(pVm/RT)=(12.315× 1× 32/0.0821× 300)=16g O2 to be released =32-16=16g

Q. A 1 L flask contains 32 g $O_{2}$ gas at $27^{\circ} C$ . What mass of $O_{2}$ must be released to reduce the pressure in the flask to 12.315 atm?

8 g

16 g

24 g

32 g

$p V = n RT = \frac{w}{m} Rt$ $w = \frac{p Vm}{RT} = \frac{12.315 \times 1 \times 32}{0.0821 \times 300} = 16 g$ $O_{2}$ to be released $= 32 - 16 = 16 g$

Q. A 1 L flask contains 32 g O2​ gas at 27∘C . What mass of O2​ must be released to reduce the pressure in the flask to 12.315 atm?

8 g

16 g

24 g

32 g

pV=nRT=mw​Rt w=RTpVm​=0.0821×30012.315×1×32​=16g O2​ to be released =32−16=16g

Q. A 1 L flask contains 32 g $O_{2}$ gas at $27^{\circ} C$ . What mass of $O_{2}$ must be released to reduce the pressure in the flask to 12.315 atm?

$p V = n RT = \frac{w}{m} Rt$ $w = \frac{p Vm}{RT} = \frac{12.315 \times 1 \times 32}{0.0821 \times 300} = 16 g$ $O_{2}$ to be released $= 32 - 16 = 16 g$