The Kelvin temperature required for 0.0470 mol of helium gas to fill a balloon to 1.20 L under 0.878 atm is:

Question

a. 0.679 K
b. 14.4 K
c. 273 K
d. 298 K
e. 307 K

Answer 1

To solve this problem, we can use the ideal gas law equation:

PV = nRT

Where:
- P is the pressure in atm
- V is the volume in liters
- n is the number of moles
- R is the ideal gas constant (0.0821 L.atm/mol.K)
- T is the temperature in Kelvin

First, we need to convert the pressure to atm:

0.878 atm = 0.878 atm

Then, we rearrange the ideal gas law equation to solve for T:

T = (PV) / (nR)

T = (0.878 atm * 1.20 L) / (0.0470 mol * 0.0821 L.atm/mol.K)
T = 22.72 / 3.839 K
T = 5.91 K

Therefore, the Kelvin temperature required for 0.0470 mol of helium gas to fill a balloon to 1.20 L under 0.878 atm is approximately 5.91 K. The closest option is:

b. 14.4 K