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General Gas Law in Physics

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?

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)

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?

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?

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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