Question

The pressure inside the cabin of an airplane decreases to 12.5 psi during a high altitude flight. Convert this pressure value to units of …. (1 psi= 0.068 atm: This is another common pressure)

a. ... millimeters of mercury (mm Hg):
b. Submit
c. ... atmospheres (atm):
d. Submit
e. ... kilopascals (kPa):

The new helium gas cylinders have just arrived at the loading dock of a local party store. In the cool January air, the 20.0-L cylinders have a pressure of 55.7 atm. Once brought into the store, the cylinder’s temperatures increases to 31.9°C and the pressure adjusts to 61.3 atm. What is the outdoor temperature in °C? (Assume a constant volume.)

A sample of oxygen gas at 68.5°C and 3.88 atm is heated until its pressure has increased to 6.10 atm. What is the new temperature (in °C)?

Will B. Sellabraten purchases a set of Helium balloons at the local card shop. Each balloon occupies a volume of 15.8 liters when in the store; but when taken outside, they are observed to shrink. If the indoor temperature is 26.6°C and the outside air temperature is -2.4°C, then what is the new volume of the balloons when taken outside?

A sample of helium gas at 61.8°C has a volume of 342 mL. What is the volume of this gas at -71.7°C?

Answer 1

Step-by-step explanation:

a. To convert 12.5 psi to mm Hg:

12.5 psi x (760 mm Hg / 14.696 psi) = 650.9 mm Hg

Therefore, the pressure inside the cabin of the airplane is 650.9 mm Hg.

c. To convert 12.5 psi to atm:

12.5 psi x (1 atm / 14.696 psi) = 0.850 atm

Therefore, the pressure inside the cabin of the airplane is 0.850 atm.

The solution for part b is missing.

To solve part d, we can use the combined gas law:

(P1 x V1 / T1) = (P2 x V2 / T2)

Where P is pressure, V is volume, and T is temperature. Assuming a constant volume, we can simplify this equation to:

P1 / T1 = P2 / T2

Solving for the new temperature (T2):

T2 = (P2 x T1) / P1

T2 = (6.10 atm x 341.65 K) / 3.88 atm

T2 = 538.4 K

Converting to Celsius:

T2 = 265.3°C

Therefore, the new temperature of the oxygen gas is 265.3°C.

To solve part e, we can use the following equation:

V2 = V1 x (T2 / T1)

Where V is volume and T is temperature. Substituting the given values:

V2 = 15.8 L x (270.8 K / 299 K)

V2 = 14.3 L

Therefore, the new volume of the helium balloons when taken outside is 14.3 L.

To solve part f, we can use the following equation:

V2 = V1 x (T2 / T1)

Where V is volume and T is temperature. Converting the temperatures to Kelvin:

T1 = 61.8°C + 273.15 = 335.95 K

T2 = -71.7°C + 273.15 = 201.45 K

Substituting the given values:

V2 = 342 mL x (201.45 K / 335.95 K)

V2 = 205.7 mL

Therefore, the volume of the helium gas at -71.7°C is 205.7 mL.