To solve this problem, we can use the ideal gas law equation, which is PV = nRT.

Here,
P = pressure = 105 kPa = 105,000 Pa (since 1 kPa = 1000 Pa)
n = number of moles = 0.153 mol
R = ideal gas constant = 8.314 J/(mol.K)
T = temperature = 28°C = 301.15 K (since Kelvin = Celsius + 273.15)

We need to find V, the volume.

Rearranging the ideal gas law equation to solve for V, we get V = nRT/P.

Substituting the given values into the equation, we get:

V = (0.153 mol * 8.314 J/(mol.K) * 301.15 K) / 105,000 Pa

Solving this gives us V = 0.0369 m³.

Since 1 m³ = 1000 L, we convert the volume to litres by multiplying by 1000.

V = 0.0369 m³ * 1000 = 36.9 L.

So, the volume occupied by 0.153 mol of oxygen gas at a pressure of 105 kPa and a temperature of 28 oC is approximately 36.9 L.

However, this answer does not match any of the options provided in the question. Please check the question again.

Question

To solve this problem, we can use the ideal gas law equation, which is PV = nRT.

Here, 
P = pressure = 105 kPa = 105,000 Pa (since 1 kPa = 1000 Pa)
n = number of moles = 0.153 mol
R = ideal gas constant = 8.314 J/(mol.K)
T = temperature = 28°C = 301.15 K (since Kelvin = Celsius + 273.15)

We need to find V, the volume.

Rearranging the ideal gas law equation to solve for V, we get V = nRT/P.

Substituting the given values into the equation, we get:

V = (0.153 mol * 8.314 J/(mol.K) * 301.15 K) / 105,000 Pa

Solving this gives us V = 0.0369 m³.

Since 1 m³ = 1000 L, we convert the volume to litres by multiplying by 1000.

V = 0.0369 m³ * 1000 = 36.9 L.

So, the volume occupied by 0.153 mol of oxygen gas at a pressure of 105 kPa and a temperature of 28 oC is approximately 36.9 L.

However, this answer does not match any of the options provided in the question. Please check the question again.

Knowee AI · Accepted Answer

To solve this problem, we can use the ideal gas law equation, which is PV = nRT.

Here, 
P = pressure = 105 kPa = 105,000 Pa (since 1 kPa = 1000 Pa)
n = number of moles = 0.153 mol
R = ideal gas constant = 8.314 J/(mol.K)
T = temperature = 28°C = 301.15 K (since Kelvin = Celsius + 273.15)

We need to find V, the volume.

Rearranging the ideal gas law equation to solve for V, we get V = nRT/P.

Substituting the given values into the equation, we get:

V = (0.153 mol * 8.314 J/(mol.K) * 301.15 K) / 105,000 Pa

Solving this gives us V = 0.0369 m³.

Since 1 m³ = 1000 L, we convert the volume to litres by multiplying by 1000.

V = 0.0369 m³ * 1000 = 36.9 L.

So, the volume occupied by 0.153 mol of oxygen gas at a pressure of 105 kPa and a temperature of 28 oC is approximately 36.9 L.

However, this answer does not match any of the options provided in the question. Please check the question again.

Calculate the volume (in litres) occupied by 0.153 mol of oxygen gas at a pressure of 105 kPa and a temperature of 28 oC.Question 23Answer383 L0.339 L23.8 L3.65 L0.114 L

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