Calculate the Fall in temperature of helium initially at 15 C when it is suddenly expanded to 8 times into volume.
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Answered by
25
Data:
Celsius temperature converted into kelvin
15 +273 = 288.1 kelvin
288.1kelvin = t1
T2=?
V1=x
V2=8x
Formula:
V1t1=v2t2
X * 288.1 = 8x * t2
x/8x * 288.1 = t2
36.01 kelvin= t2
-237.14 celsius=t2
Answer
Fall in temperature of helium is -237.14 celsius when volume expanded 8 times.
Answered by
21
Answer:
answer is above.... hope that helps you..
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