A GAS TAKES UP 25.2 L AT 25*C. AT 25*c THE GAS CAN ALSO TAKE UP 12.2lAT 1520 TORR. WHAT WAS THE PRESSURE IN ATM, OF THE ORIGINAL SAMPLE? I GOT 4.1S ATM. IS THAT RIGHT?
I don't know what an answer of 4S means. I would use P1V1 = P2V2 with P in torr, then convert to atm. My answer is about 0.9 atm
Very good answer
To solve this problem, we can use the combined gas law formula, which relates the initial and final volumes and pressures of a gas:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 and P2 are the initial and final pressures of the gas
V1 and V2 are the initial and final volumes of the gas
T1 and T2 are the initial and final temperatures of the gas (in Kelvin)
Let's convert the temperatures to Kelvin:
Given T1 = 25 °C, we can convert it to Kelvin by adding 273.15 to get T1 = 298.15 K.
Given T2 = 25 °C, we convert it to Kelvin as well to get T2 = 298.15 K.
Now we can substitute the given values into the formula:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
(unknown pressure in atm * 25.2 L) / 298.15 K = (1520 Torr * 12.2 L) / 298.15 K
Now, let's rearrange the equation to solve for the unknown pressure in atm:
(unknown pressure in atm) = [(1520 Torr * 12.2 L) / (298.15 K)] * (298.15 K / 25.2 L)
Now we can simplify:
(unknown pressure in atm) = (19440 Torr L) / (25.2 L)
(unknown pressure in atm) = 772.86 Torr
Finally, let's convert Torr to atm:
(unknown pressure in atm) = 772.86 Torr * (1 atm / 760 Torr)
(unknown pressure in atm) = 1.017 atm
Therefore, the pressure in atm of the original sample is approximately 1.02 atm (rounded to two decimal places).
So, the answer 4.1 s atm is not correct. The correct answer is approximately 1.02 atm.