a gas occupied 500.0L when it was collected over water at a pressure of 750mmHg and a temperature of 21 C. calculate the volume of dry gas at 730mmHg and 5 C. It would be a very big help if someone can help me answer this. Please and thank you!
you only need to apply the ideal gas law, PV=nRT. You're comparing two conditions of the same system, so you can move all the terms to one side, \(\dfrac{PV}{nRT}=C\) and make it equal to the other system. so it's \(\underbrace{\dfrac{P_1V_1}{nRT_1}}=\underbrace{\dfrac{P_2V_2}{nRT_2}}\) conditions 1 conditions 2 since since it's a closed system, n is constant (R is always constant so we can drop them. the formula is reduced to: \(\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\)
Okay. and we are looking for V2 correct? I've just been very very confused with this question. I appreciate you helping me!
no problem ! and yes you're looking for the new volume, \(V_2\). So convert the temp to the Kelvin scale, plug in your values, then solve (algebraically) for the new V.
okay, this is what I did. V2=(T2)(P1)(V1)/(T1)(P2)
V2= (278K)(750mmHg)(500.0L)/(294K)(730mmHg)
V2= 258852040.82 but something seems weird. this answer is too large lol
i got 485.7422421023203802
you messed up somewhere plugging into your calculator
hahahaha!!! I'm sure I did!!!
haha you should do it in steps, first do the top multiplication, the bottom multiplication, then the top and the bottom division.
oh okay! yes I got the same answer now!
I have my answer in least number of sig figs, so would the final answer be 485.7?
I have to have my answer in*
hm so is the least number of sig figs 1, because 5 Celsius?
or are we using the sig figs of the volume 500.0 L?
well since I had to change the temperatures to kelvin... I think it would be 2 because 500.0 is 4, the temperatures are both 3, and both 750mmHg & 730mmHg are 2 sig figs
Oh I see what you are saying, going by the least number of decimal points.
hmm i would try 486 first
I'll put 486 as my final answer. Thank you so much for all of your help! I really appreciate it!!!!
no problem !
Join our real-time social learning platform and learn together with your friends!