Question

9. Consider the following apparatus. Calculate the partial pressure of helium after the opening valve. The
temperature remains constant at 16°C.
He
Ne
1.2 L
0.63 atm
3.4L
2.8 atm
c) 0.328 atm
a) 0.164 atm
b) 1.64 atm
d) 1 aatm
please tell me the answer. I will mark brainliest

Answer 1

Answer:

Total moles remains conserved:

$n_{Total} =n_{Helium} +n_{neon}$

$\frac{PV_{Total} }{RT_{1} }$=$\frac{P_{He} V_{1} }{RT_{1} }$+$\frac{P_{Ne} V_{2} }{RT_{1} }$

P(1.2+3.4)=1.2*0.63+3.4*2.8

P=$\frac{0.756+9.52}{4.6} =2.23atm$

$P_{He} =\frac{n_{He} }{n_{Total} } *2.23$

$=\frac{\frac{P_{He}V_{1}  }{RT_{1} } }{\frac{P_{Total}*V_{Total}  }{RT_{1} } }$*2.23

$=\frac{0.63*1.2}{2.23*4.6} *2.23$

=0.164atm

Answer 2

Answer:

I guess the simplest way of solving is by using Boyles law.

Original Pressure and Volume

P1=0.63atm

V1=1.2L

Final Pressure and Volume

V2=(1.2+3.4)L=4.6L

P2=?

By Boyle's law,

P1V1=P2V2

Therefore P2=(P1V1)÷V2

=O.63×1.2÷4.6

=0.164amu

Hence partial pressure of He is 0.164amu.

Explanation:

Partial Pressure is the pressure of one gas in a mixture of gases. In this question,it is easier to use Boyle's law rather than ideal gas law, since you already have the PV constant of the Helium gas. The final volume is the volume of entire vessel since gas occupies complete volume of vessel.