Asked by Lauren
If 456 dm^3 of krypton at 101 kPa and 21 degrees C is compressed into a 27.0 dm^3 tank at the same temperature, what is the pressure of krypton in the tank?
V1= 456 dm^3
V2= 27.0 dm^3
P1= 101 kPa
P2= ?
P2= 101 kPa * 456 dm^3/ 27.0 dm^3
ANSWER:
1705.78 kPa
V1= 456 dm^3
V2= 27.0 dm^3
P1= 101 kPa
P2= ?
P2= 101 kPa * 456 dm^3/ 27.0 dm^3
ANSWER:
1705.78 kPa
Answers
Answered by
Lauren
Is this right??
Answered by
DrBob222
Almost.
All of the digits are correct because the answer is a repeating number of 1705.7777777. However, you have only three (3) significant figures in the problem; therefore, you may have only 3 s.f. in the answer. So the answer should be rounded to 1.70 x 10^3. In rounding, and when the last number is a 5, I round to the nearest even number so the 0 stays and we drop the 5. Had the zero been a 1, we would have rounded up by making the 1 a 2 and droping the 5. And the scientific notation is necesssary in the answer because if we write it as 1700 we are showing four s.f.
All of the digits are correct because the answer is a repeating number of 1705.7777777. However, you have only three (3) significant figures in the problem; therefore, you may have only 3 s.f. in the answer. So the answer should be rounded to 1.70 x 10^3. In rounding, and when the last number is a 5, I round to the nearest even number so the 0 stays and we drop the 5. Had the zero been a 1, we would have rounded up by making the 1 a 2 and droping the 5. And the scientific notation is necesssary in the answer because if we write it as 1700 we are showing four s.f.