Hello!
If I initially have a gas at a pressure of 12 atm, volume of 23 liters, and temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?
We have the following data:
P1 (initial pressure) = 12 atm
V1 (initial volume) = 23 L
T1 (initial temperature) = 200 K
P2 (final pressure) = 14 atm
T2 (final temperature) = 300 K
V2 (final volume) = ? (in L)
<u><em>Now, we apply the data of the variables above to the General Equation of Gases, let's see:</em></u>


multiply the means by the extremes




<u><em>*** Note: the approximation rule says that when the number before the digit 5 is odd, the previous value is raised to the next even number</em></u>
<u><em>Answer:</em></u>
<u><em>The new volume is approximately 30 Liters </em></u>
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