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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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