A gas at one atmosphere and 100 mL is mixed with another gas of equal moles at 0.5 atm and having volume 50 mL in flask of 1 L. What is their final pressure?

Question

A gas at one atmosphere and 100 mL is mixed with another gas of equal moles at 0.5 atm and having volume 50 mL in flask of 1 L. What is their final pressure? (A) 0.5 atm (B) 1 atm (C) 0.75 atm (D) 0.125 atm

Tardigrade Admin · Accepted Answer

No. of moles in mixture is the sum of separate no. of moles of gases. n = n1 + n2 or (PV/RT) = (P1V1/RT)+(P2V2/RT) or P Ã1000 = 1 Ã 100 + 0.5 Ã50 or P = (125/1000) atm = 0.125 atm

Q. A gas at one atmosphere and 100 mL is mixed with another gas of equal moles at 0.5 atm and having volume 50 mL in flask of 1 L. What is their final pressure?

0.5 atm

1 atm

0.75 atm

0.125 atm

No. of moles in mixture is the sum of separate no. of moles of gases.
$n = n_{1} + n_{2}$
or $\frac{P V}{RT} = \frac{P _{1} V _{1}}{RT} + \frac{P _{2} V _{2}}{RT} or P \times 1000 = 1 \times 100 + 0.5 \times 50$
or $P = \frac{125}{1000} a t m = 0.125 a t m$

Q. A gas at one atmosphere and 100 mL is mixed with another gas of equal moles at 0.5 atm and having volume 50 mL in flask of 1 L. What is their final pressure?

0.5 atm

1 atm

0.75 atm

0.125 atm

No. of moles in mixture is the sum of separate no. of moles of gases. n=n1​+n2​ or RTPV​=RTP1​V1​​+RTP2​V2​​orP×1000=1×100+0.5×50 or P=1000125​atm=0.125atm

No. of moles in mixture is the sum of separate no. of moles of gases.
$n = n_{1} + n_{2}$
or $\frac{P V}{RT} = \frac{P _{1} V _{1}}{RT} + \frac{P _{2} V _{2}}{RT} or P \times 1000 = 1 \times 100 + 0.5 \times 50$
or $P = \frac{125}{1000} a t m = 0.125 a t m$