An expandable balloon contains 1400L of He at .950atm pressure and 18 degrees Celcius. At an altitude of 22 miles (temperature of 2.0 degrees Celcius and pressure of 4.0 torr), what will be the volume of the balloon?
P*V/T = constant if T is in K
0.95*1400/291 = (4/760)*V/275
V is the final volume, at altitude. Solve fore that.
Note that I had to use the same units for P, V and T on both sides of the equation. 4 torr = 4/760 atm
To solve this problem, we can use the ideal gas law, which states:
PV = nRT
where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
First, let's convert the given temperatures to Kelvin:
Initial temperature (T1) = 18 °C = 18 + 273.15 = 291.15 K
Final temperature (T2) = 2.0 °C = 2.0 + 273.15 = 275.15 K
Next, let's convert the given pressures to atm:
Initial pressure (P1) = 0.950 atm
Final pressure (P2) = 4.0 torr = 4.0 / 760 atm (since 1 atm = 760 torr)
Now, we need to find the number of moles (n). We can use the ideal gas law to solve for n:
n = (P * V) / (R * T)
Since the amount of helium (He) is constant (assuming no leaks or chemical reactions), we can set the initial and final number of moles equal to each other:
(P1 * V1) / (R * T1) = (P2 * V2) / (R * T2)
Rearranging the equation and solving for V2, we get:
V2 = (P1 * V1 * T2) / (P2 * T1)
Now, let's plug in the given values:
P1 = 0.950 atm
V1 = 1400 L
T1 = 291.15 K
P2 = 4.0 / 760 atm
T2 = 275.15 K
Calculating V2:
V2 = (0.950 * 1400 * 275.15) / (4.0 / 760 * 291.15)
= (371570.6) / (1.09)
≈ 341009.4 L
Therefore, the volume of the balloon at the altitude of 22 miles will be approximately 341009.4 liters.
To solve this problem, we can use the ideal gas law equation, which states:
PV = nRT
Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, let's convert the given values to the appropriate units:
Initial conditions:
Initial pressure (P1) = 0.950 atm
Initial volume (V1) = 1400 L
Initial temperature (T1) = 18 °C = 18 + 273.15 K = 291.15 K
Final conditions:
Final pressure (P2) = 4.0 torr = 4.0/760 atm = 0.00526 atm
Final temperature (T2) = 2 °C = 2 + 273.15 K = 275.15 K
Now, we can solve for the final volume (V2) using the ideal gas law equation:
P1V1/T1 = P2V2/T2
Substituting the given values:
(0.950 atm)(1400 L)/(291.15 K) = (0.00526 atm)(V2)/(275.15 K)
Simplifying the equation:
(0.950)(1400) = (0.00526)(V2)
Dividing both sides by (0.00526):
(0.950)(1400)/(0.00526) = V2
V2 ≈ 252,626 L
Therefore, at an altitude of 22 miles, the volume of the balloon is approximately 252,626 liters.