Chemistry Forum

Hey guys, I need help with this question. Super confused on it. Here it is:

A balloon is inflated to 4.50 L with Helium, at a particular initial temperature and pressure. (It is tied off.) Now, assume that the temperatures and pressures referred to are always those prevailing inside the balloon.

Then, what would the volume of the balloon be when the pressure is reduced to 1/3 of the initial value and the Kelvin temperature is reduced to 1/2 of the initial value?

I have a feeling of where to start but what are the values of the initial pressure? That part is throwing me off!

If anyone could help, that would be nice.

Thanks

Do you know the ideal gas law equation pV = nRT
So you solve for V
Change now p to1/3p and T to 1/2T

Oh wow, that makes sense! Thank you!

So, what is your final answer? What is the ratio of temperature to pressure initially and after making adjustments to their values?

I got 0.0250 L. But that was incorrect... not sure what I'm doing wrong?

Initially, there was 4.5 L volume of Helium gas in the Balloon. So, what is your computed number of mole of Helium gas?

4.50 L x 1 mol He / 22.414 L = 0.200767377 mol (not rounded up)

So, Plugging all the available values in to the formula PV=NRT, we get the ratio of Kelvin temperature to pressure in Bar as 2.696927647572174. Thereafter the new ratio is \(\frac{X}{2} \times \frac{3}{Y}\rightarrow \frac{3X}{2Y}\) where X=Kelvin Temperature , Y= Pressure in Bar.

What are the next computational workings you made to arrive at the final answer of 0.0250 L volume of Helium Gas?

Hmm.. I just inserted all the values into the equation V=nRT/P. I think I made a mistake though because I got a different answer to the one I wrote before. My new answer is 0.02746 L. Still wrong..

By the way, how did you get that equation? How did you come up with that thinking?

Question itself states my thinking. Question says one-half of initial value of temperature(X) and one third of initial value of pressure(Y). So ratio becomes \(\frac{\frac{X}{2}}{\frac{Y}{3}}\).Simplifying, we get \(\frac{3X}{2Y}\). That's it.

The result is 1.5 x V.

That's correct. thanks. If, i am not wrong, the final volume is 4.04 L of Helium gas in the balloon.

You guys have been a real help! Thank you

Its wrong. The result is 6.75 l

Oh, Yes. You are correct. 4.04539 is the new ratio of Kelvin temperature to pressure after making adjustments to their initial values. Thanks for pointing out my mistake.