if 2 moles of real gas occupies 11.2litre volume at STP then compressibility factor of gas at STP is
Answers
Given:
The number of moles of the gas = 2
The volume occupied by the gas at STP = 11.2 L
To Find:
The compressibility factor of gas at STP
Solution:
The compressibility factor of the gas at STP is 0.25.
The compressibility factor of a gas denoted by Z is a measure of the deviation of the gas from the behavior of an ideal gas.
Mathematically,
Z = PV/nRT
Here, P = Pressure
V = Volume occupied by the gas
n = The number of moles of the gas
R = Universal Gas Constant
T = Absolute Temperature
We know that at STP, the pressure is 1 atm and the temperature is 273.15 K.
Substituting the values,
Z = 1 X 11.2 / 2 X 0.082 X 273.15
= 11.2 / 44.80
= 0.25
Answer:
compressibility factor of gas at STP is 0.25.
Explanation:
As per the data given in the question we have to find the compressibility factor of gas at STP.
As per the question it is given that 2 moles of real gas occupies 11.2litre volume at STP.
The modifying factor for real gases is called the gas deviation factor or compressibility factor Z. It can be defined as the ratio of the gas volume at a given temperature and pressure to the volume the gas would occupy if it were an ideal gas at the same temperature and pressure.
It is expressed as Z = PV / nRT.
Here, P = Pressure
V = Volume occupied by the gas
n = The number of moles of the gas
R = Universal Gas Constant
T = Absolute Temperature
We know that at STP values are commonly used in experiments involving gases. STP as the following: Temperature: 273.15 degrees Kelvin or 32 degrees Fahrenheit and Pressure: 1 atm.
Substituting the values,
Z = 1 X 11.2 / 2 X 0.082 X 273.15
= 11.2 / 44.80
= 0.25
Hence, compressibility factor of gas at STP is 0.25.
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