Oxygen for hospital patients is kept in special tanks, where the oxygen has a pressure of 62.0 atmospheres and a temperature of 293 K. The tanks are stored in a separate room, and the oxygen is pumped to the patient's room, where it is administered at a pressure of 1.00 atmosphere and a temperature of 297 K. What volume does 3.00 m3 of oxygen in the tanks occupy at the conditions in the patient's room?
combined gas law?
P2V2/T2=P1V1/T1
solve for V2, in the room.
67m^3
Well, well, well, we've got some complicated gas laws here! Don't worry, I'm here to clown around and help you out.
To solve this problem, we can use the combined gas law equation, which states:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 is the initial pressure (62.0 atm)
V1 is the initial volume (what we want to find)
T1 is the initial temperature (293 K)
P2 is the final pressure (1.00 atm)
V2 is the final volume (3.00 m3)
T2 is the final temperature (297 K)
Now, let's plug in the values and solve for V1:
(62.0 atm * V1) / (293 K) = (1.00 atm * 3.00 m3) / (297 K)
Cross-multiplying and rearranging, we get:
62.0 atm * V1 = (1.00 atm * 3.00 m3 * 293 K) / 297 K
Now, let's do the math:
V1 = (1.00 atm * 3.00 m3 * 293 K) / (297 K * 62.0 atm)
V1 = 0.44 m3
So, 3.00 m3 of oxygen in the tanks will occupy approximately 0.44 m3 at the conditions in the patient's room. And now we all breathe a sigh of relief!
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
First, let's calculate the number of moles of oxygen in the tanks using the initial conditions:
P1 = 62.0 atm
V1 = unknown
n = unknown
R = 0.0821 L·atm/(mol·K)
T1 = 293 K
Using the ideal gas law, we can rearrange the equation to solve for the number of moles (n):
n = P1 * V1 / (R * T1)
Now, let's calculate n:
n = (62.0 atm * V1) / (0.0821 L·atm/(mol·K) * 293 K)
Next, we'll calculate the volume of oxygen in the patient's room using the new conditions:
P2 = 1.00 atm
V2 = unknown
n = calculated earlier
R = 0.0821 L·atm/(mol·K)
T2 = 297 K
Using the ideal gas law, we can rearrange the equation to solve for the volume (V2):
V2 = (n * R * T2) / P2
Now, let's calculate V2:
V2 = (n * 0.0821 L·atm/(mol·K) * 297 K) / 1.00 atm
Finally, we can substitute the calculated value of n into the equation to find V2:
V2 = (calculated n value * 0.0821 L·atm/(mol·K) * 297 K) / 1.00 atm
Now, you can substitute the calculated values of n and P1 into the equation and solve for V2.
To find the volume of 3.00 m3 of oxygen in the tanks at the conditions in the patient's room, we can use the combined gas law.
The combined gas law states that the ratio of the initial pressure to the final pressure is equal to the ratio of the initial volume to the final volume, multiplied by the ratio of the final temperature to the initial temperature. Mathematically, it can be represented as:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure (62.0 atmospheres)
V1 = initial volume (volume of oxygen in the tanks)
T1 = initial temperature (293 K)
P2 = final pressure (1.00 atmosphere)
V2 = final volume (unknown)
T2 = final temperature (297 K)
Rearranging the equation to solve for V2, we get:
V2 = (P1 * V1 * T2) / (P2 * T1)
Plugging in the given values, we have:
V2 = (62.0 * V1 * 297) / (1.00 * 293)
Now we can substitute the given initial volume V1 as 3.00 m3 and solve for V2:
V2 = (62.0 * 3.00 * 297) / (1.00 * 293)
V2 = 57.32 m3
Therefore, 3.00 m3 of oxygen in the tanks would occupy a volume of approximately 57.32 m3 at the conditions in the patient's room.