Introduction
The general gas law in physics is the combination of Boyle’s, Charle’s, and Pressure laws for an ideal gas.
We can write the formula as
PV / T = Constant
Where
P = pressure
V = volume
T = Temperature
And because of the constant, we can now write the formula as
P1V1 / T1 = P2V2 / T2
General Gas Law in Physics Practice Problems
Here are a few examples to help you understand how to apply the formula for general gas law and calculate a problem.
Problem 1
A mass of gas at 170c and 700mmHg has a volume of 1.2m3. Determine its volume at 270c and pressure at 750 mmHg?
Answer
Solution: To calculate the volume from general gas law, we need to first extract our data from the question
Data:
Initial Pressure = P1 = 700mmHg
Final Pressure = P2 = 700mmHg
Initial Temperature = T1 = 170c = 273 + 17 = 290
Final Temperature = T2 = 270c = 273 + 27 = 300
Additionally, initial Volume = V1 = 1.2m3
Also, Initial Volume = V1 = ?
Using the general gas law,
( P1 V1) / ( T1 )= ( P2 V2 ) / ( T2 )
By Making V2 subject of the formula,we have
Thus, V2 = ( P1 V1T2) / (P2 T1 )
by substituting the above expression with our data, we have:
Final volume, V2 = ( 700 x 1.2 x 300 ) / ( 750 x 290 )
Therefore, V2 = ( 252,000 ) / ( 217,500 )
This will give us our final answer as V2 = 1.156m3
Therefore, the final volume is 1.2m3
Problem 2
The volume and pressure of a given mass of gas at 270C are 76 cm3 and 80 cm of mercury respectively. Calculate its volume at standard temperature and pressure (s.t.p)
Answer
Data
The original volume, V1 = 76 cm3
Initial pressure, P1 = 80 cmHg
Final pressure, P2 = 760 cmHg [Because of the standard temperature and pressure]
The initial temperature, T1 = 270C = 27 + 273 = 300 k
Final temperature, T2 = 273k [Because of the standard temperature and pressure]
Unknown
The final volume, V2 = ?
Formula
We will now apply our formula to solve the problem
( P1 V1) / ( T1 )= ( P2 V2 ) / ( T2 )
after inserting our data, we will now have
(80 x 76) / (300) = (76 x V2) / 273
By making V2 subject of the formula, we will obtain our answer as
V2 = 72.8 cm3
Problem 3
A given mass of an ideal gas occupies a volume V at a temperature T and under a pressure P. If the pressure is increased to 2P and the temperature reduced to (1/2)T, then what is the percentage change in the volume of the gas?
Answer
The final answer to the above question is 75%
Explanation
Since we have a formula that says P1V1 / T1 = P2V2 / T2
and
V1 = V
P1 = P
P2 = 2P
T1 = T
T2 = (1/2)T
V2 = ?
We can make V2 subject of the formula and interpret it as
V2 = (P x V x (1/2)T ) / (2P x T)
And we will have
V2 = (1/2)V / 2 = 0.25 V
The percentage change in volume will now become
[(V1 – V2) / V1 ] x 100%
Which will now be
[(V – 0.25V) / V ] x 100%
hence
[0.75V / V ] x 100% = 0.75 x 100% = 75%
Therefore, the percentage change in the volume of the gas is 75%
Problem 4
The pressure of a given mass of a gas changes from 300 Nm-2 to 120 Nm-2 while the temperature drops from 1270C to -730C. What is the ratio of the final volume to the initial volume?
Answer
The final answer to the question is V1 : V2 = 5 : 4
Explanation
P1 = 300 Nm-2
and P2 = 120 Nm-2
T1 = 127 + 273 = 400 K
we have T2 = -73 + 273 = 200 K
Formula
The formula we will use is P1V1 / T1 = P2V2 / T2
Solution
We will now insert our data into the formula
(300 x V1) / 400 = (120 x V2) / 200
To have a ratio of V1 to V2
Hence,
V2 : V1 = 60000 / 48000
60 / 48 = V2 / V1
When we divide both 60 and 48 by 12, we will obtain
V2 / V1 = 5 / 4
You may also like to read:
How to Calculate Cubic Expansivity with Examples
Linear Expansivity: Definition and Calculations
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