For every 3 deg rise in temperature, the volume of a certain gas expands by 4 cubic centimeters. If the volume of the gas is 24 cubic centimeters when the temperature is 32 deg, what was the volume of the gas in cubic centimeters when the temperature was 20 deg??
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I know that answer, just help me figure it out plsplsplsplsplpslpslpslpslplsplsplsplspls...
I just went backwards...
T=32-3=29 so volume at that time is=24-4=20
T=29-3=26 , v=16
T=20, v =8 (answer)
It didn't take time so it could be a good strategy for this specific question.
Using arthimetic sequence(Really isn't needed and takes a lot of time for this specific question but good if the numbers were far from each other)
We don't know the current temperature (Start)
So let Temperature= T
To find T , we know that at somepoint the temperature was 32 and volume was 24
Where did the 3 come from? "For every 3 deg rise in Temp" So going backward from 32 degrees We wouldn't add 3. We subtract then
L is a specific number of terms.. For example, If 32 was for example, the 5th term and i want the 2nd term
therefore 32-3 -3 -3 etc..
Also could be said as: 32-3L
So Now we know that temperature is equal to , T=32-3L
Same thing goes for the volume.
V=The point which was given with T (t=32 degrees, v=24 cubic cen v) -4L
where did the 4 come from?
for every 3 degrees, the volume expands by 4.
Going backwards, for decreasing 3 degrees, the volume shrinks(Decreases) by 4
L no. of terms backwards as well.
We now know that:
"What was the volume of the gas in cubic cent. when the temperature was 20 degrees?"
So T=32-3L=20 (deg)
Solve the equation for L
Subsituite L=4 in Volume equation
V=24-4(4) = 8
Again, the strategy I used earlier above would save a lot of time but who knows what they would put in the exam.. It wouldn't work for far numbers apart. So Using arithmetic sequence is better.
There might be a better answer. That's just my answer.
Hope that helped :).