Ideal Gas Laws

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ChenBeier
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Re: Ideal Gas Laws

Post by ChenBeier »

Where do you get this number.
The balloon has 4.5 l first now we decrease pressure to 1/3 p. If everything is constant then the volume will expand to 3 x 4,5 l = 13.5 l
Now we decrease the temperatur to half, the volume decrease also to half what means 6.75 l
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Dhamnekar Winod
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Re: Ideal Gas Laws

Post by Dhamnekar Winod »

I am showing you my workings on this question.
PV= nRT where T= xKelvin and P= yBar
yBar* 4.5 L = 0.2007 mol *8.314472 J/(mol* K) * xK \(\Rightarrow \frac{4.5 L}{1.6686}=\frac{xJ}{yBar} \rightarrow \frac{x}{y} L\)

So, \(\frac{x}{y}= 2.6969\) The new ratio is \(\frac{3x}{2y}\rightarrow 2.6969*1.5=4.0454\)

So, the ratio of temperature x(unitless) to pressure y (unitless) is 4.0454. From here, we can easily compute final volume of Helium gas in the balloon which is 6.75 L as you computed.\(\frac{4.0454 * 4,5 L}{2.6969}= 6.75 L\)
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