Themodynamics: calculating the partial pressure with Dalton's Law

[erde]

Homework Statement

In a sealed container is Helium $M_{He} = \frac {4kg} {kmol}$ with a pressure of $p_{He} = 4bar$. now is Methan put isothermic inside the container till both the methan and the helium mass are equal( $M_{CH4} = \frac {16kg} {kmol}$ Calculate using the ideal gas law and Dalton´s law the total pressure $p_{total}$ after the procedure

Homework Equations

$P*V=m*R*T$
ψ: volumen %

The Attempt at a Solution

[/B]
i have two different attempts with two different results, can someone please help me to identify which one is correct?
attempt 1.
1) $P_{He} V_{He} = m_{He} R_{He} T_{He}$
2) $P_{CH4} V_{CH4}= m_{CH4} R_{CH4} T_{CH4}$
$\frac {1)} {2)} = \frac {P_{He} V_{He}} {P_{CH4} V_{CH4}}= \frac {R_{He}} {R_{CH4}}$

3) $R=\frac {R_m} {M_M}$

$\frac {1)} {2)} = \frac {P_{He} V_{He}} {P_{CH4} V_{CH4}} = \frac {M_{CH4}} {M_{He}} = \frac {16} {4} =4$

$\frac {V_{He}} {V_{CH4}}=\frac {V_{He}} {V_{total}} \frac {V_{total}}{V_{CH4}} = \frac {ψ_{He}} {ψ_{CH4}} =\frac {p_{He}}{p_{CH4}}$

$\frac {p_{He} p_{He}}{p_{CH4} p_{CH4}}=4$
$16bar^2= p_{He} ^2= 4 p_{CH4} ^2$
$4bar^2= p_{CH4} ^2$
$p_{CH4}= 2 bar$
Dalton: $p_{CH4} + p_{He}=p_{total}=6bar$attempt 2.
AVOGADRO 1mol of gas at 25 °C occupies 24l so:
n: mol;
$R_m$: mol gas konstant;
$dT=0$ for isothermic reaction $T=constant$
$pV=nR_mT$
if $p_{He}$ is 4 bar and we assume that the Volumen of the container is $24L$ and the temperature is $25°C$ the should be 4 mol of Helium in the container following the equation
$p_{1bar} V_{24l}= n_{1mol} R_m T$
$p_{4bar} V_{24l}= n_{4mol} R_m T$
$V$, $R_m$ and $T$ are constant so increasing $n$ from $1$ to $4$ should also increase $p$ from $1$ to $4$

so if we input Methan till $m_{Methan}=m_{Helium}$ and we now that methan weights 4 times the weight of helium per mol we now that there are $1 mol_{Methan}$ for every $4mol_{Helium}$.
so if the total mol amount inside the container is 5 mol the total pressure should be 5 bar
$p_{5bar} V_{24l}= n_{5mol} R_m T$
so $p_{total}=5bar$

 

[DrClaude]

Neither approach is very good. The first one is incorrect because you do not take into account the fact that since the total volume does not change, then the volume of helium is not the same before and after the methane is added.

In the second case, you shouldn't assume values of the volume and the temperature. You should simply work with a volume $V$ and a temperature $T$.

 

[erde]

so for the second approach the correct way would be to say
1)$p_0 V_0= n_0 R_m T_0$
2)$4p_0 V_0= 4n_0 R_m T_0$
$\frac {2)}{1)}=\frac {4p_0}{p_0}=\frac {4n_0}{n_0}$
so if $m_{CH4}=m_{He}$ => $n_{CH4}=1/4n_{He}$
3) $xp_0 V_== (4+1/4*4)n_0 R_m T_=$ => $xp_0 V_0=5n_0 R_m T_0$ => $x=5$ so $p_{total}=5bar$