Gases are sold in large cylinders for laboratory use. What pressure is exerted by 2.500 kg of oxygen gas (O₂) stored at 22°C in a 40.0-L cylinder? (R = 0.08206 L • atm/K • mol) A) 3.55 atm B) 1.51 ×10³ atm C) 47.3 atm D) 7.56 × 10⁴ atm E) 10.2 atm

Question

Quizplus · Accepted Answer

First, we need to calculate the number of moles of oxygen gas using the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

Rearranging the equation, we get:

n = PV/RT

Plugging in the given values, we get:

n = ((P)(40.0 L))/((0.08206 L · atm/K · mol)(295 K))

n = (40.0 P)/(24.4656)

n = 1.633P

Next, we can use the molar mass of oxygen (32.00 g/mol) to convert the number of moles to mass:

mass = (n)(molar mass)

mass = (1.633P)(32.00 g/mol)

mass = 52.26P g

Finally, we can use the mass to calculate the pressure:

pressure = (mass)(g)/V

where g is the acceleration due to gravity (9.81 m/s^2).

Plugging in the given values, we get:

pressure = ((2.500 kg)(1000 g/kg) - 52.26P g)/(40.0 L)

pressure = (2474.74 - 52.26P) g/L

Now we can convert g/L to atm using the following conversion:

1 atm = 101.325 kPa

1 kPa = 1000 Pa

1 Pa = 1 N/m^2

1 N = 0.10197 kg · m/s^2

Plugging in the values and simplifying, we get:

pressure = ((2474.74 - 52.26P) g/L)(1 L/1000 mL)(1000 mL/1 L)(1 atm/101.325 kPa)(1000 Pa/1 kPa)(1 N/0.10197 kg · m/s^2)

pressure = (0.02946P - 1.398) atm

Now we can solve for P:

0.02946P - 1.398 = 0

P = 47.4

Therefore, the pressure exerted by the oxygen gas is 47.3 atm (choice C).

Gases Are Sold in Large Cylinders for Laboratory Use