What is the volume (in liters) of 2.5 moles of oxygen gas measured at 250C and a pressure of 104.5 kpa?

Question

Jeremiah Adams · Accepted Answer

#V=(nRT)/P# #~=# #100*L# Of course you must choose the appropriate Gas constant. Here it is #8.314*L*kPa*K^-1*mol^-1#. #"Volume"# #=(2.5*cancel(mol)xx8.314*L*cancel(kPa)*cancel(K^-1)*cancel(mol^-1)xx523*cancelK)/(104.5*cancel(kPa))# #=# #??L#

Natalie Anton · Answer

undefined To calculate the volume of a gas using the ideal gas law, we can use the formula:

$$V = \frac{nRT}{P}$$

Where:
- $V$ is the volume of the gas (in liters)
- $n$ is the number of moles of the gas
- $R$ is the ideal gas constant ($0.0821 \frac{L \cdot atm}{mol \cdot K}$ or $8.314 \frac{J}{mol \cdot K}$)
- $T$ is the temperature of the gas (in Kelvin)
- $P$ is the pressure of the gas (in kilopascals)

Given:
$n = 2.5 \text{ moles}$
$T = 25°C + 273.15 = 298.15 \text{ K}$
$P = 104.5 \text{ kPa}$

Substitute the given values into the formula:

$$V = \frac{(2.5 \text{ moles})(0.0821 \frac{L \cdot atm}{mol \cdot K})(298.15 \text{ K})}{104.5 \text{ kPa}}$$

$$V = \frac{(2.5)(0.0821)(298.15)}{104.5}$$

$$V ≈ \frac{(61.725)}{104.5}$$

$$V ≈ 0.591 \text{ L}$$

Therefore, the volume of 2.5 moles of oxygen gas at 25°C and a pressure of 104.5 kPa is approximately 0.591 liters.