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 \(\frac{1}{2T}\), then the percentage change in the volume of the gas is
0%
25%
75%
300%
400%
Explanation
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Now its so simple but technical. So just follow my lead😋🙌.
First parameters=P1, V1, T1
Second parameters= 2P,1/2T, V2
We can now use normal gas law formula
P1v1/T1=4PV2/T2
Cross multiply
P1V1*T2=4PV2*T1 . ( you know we were asked to find V2) so we are to make V2 the subject of the formula
V2=P1V1T2/4PT2( after cancelling, we will have)
V2=V/4(but remember that alphabets are = 1 . so we are to find the percentage change in the volume.
Increase in volume/ original volume *100%. ( the formula is cool😘)
So = 1-1/4÷1*100=75%
Okay watch me now.. You should understand this, if u dont understand at the end of this explanation, please a quick advice, go and wash ur head in Osun river.
Lol, just kidding.. 
Alright lets jump into it.....
Since we have Pressure(P), Volume(V) and Temperature(T) in the question, we are going to apply the general gas equation which is
P1 x V1 P2 x V2
---------- = ------------
T1 T2
Now note; every paraneters having 1, indicates initial, while the ones having 2 indicates final. For example, P1 means initial pressure while P2 means final pressure. Same for V and T. Ok good..
The question says; the mass of the gas occupies a volume V at a temperature T under a pressure P... We take this as the initial, i.e,
P1 = P; V1 = V; while T1 = T
Now the question then says; if the pressure is increased to 2P...... meaning that the initial pressure P was multiplied by 2 to give a final pressure of 2P...
The temperature was also reduced from the question to (1/2)T.... meaning that the initial pressure T was divided by 2 to give (1/2)T
{i.e: T÷2 = T/2 = (T/1) x (1/2) = (1/2)T}
(If you dont understand the explanation in the coiled bracket above, u can skip it, it is not really necessary for this calculation, just bear in mind that (1/2)T is half of T and it means
T ÷ 2, you gerrit?? if u dont gerrit.......Otilor..)
Lol.. back to work..
Now, what will be the final volume???????
Now this is where work starts, shine your eyes well well oooo.
Ok
We have initial and final pressures (P1 and P2), initial and final Temperature (T1 and T2), initial Volume(V1)... Now to get the final volume (V2), we incorporate our derived parameters from the question into our formula...
P1 = P
V1 = V
T1 = T
P2 = 2P
V2 = ????
T2 = (1/2)T
Right ????
Our formula
P1 x V1 P2 x V2
---------- = ------------
T1 T2
Becomes
P x V 2P x V2
--------- = ------------
T (1/2)T
Note: Our V2 remains V2 because that is what we are solving for.
From the above equation, we cross multiply and it becomes;
P x V x (1/2)T= 2P x V2 x T
Since we are looking for V2, we divide both sides of the equation by 2P x T, the equation becomes;
P x V x (1/2)T 2P x V2 x T
------------------- = -----------------
2P x T 2P x T
or the other way
2P x V2 x T. P x V x (1/2)T
------------------- = -----------------
2P x T 2P x T
Equation is equation, LHS is equal to RHS.
Now cancel out every term in the numerator by its fellow common term in the denominator(P will cancel out P, T will cancel out T)
What we have left is
V x (1/2)
V2 = ---------------
2
The RHS needs to be simplified.. how do we do that ??
V x (1/2) = V/2
The equation becomes,
V/2
V2 = --------
2
We can further simplify the RHS by
V/2
-------- = V/2 ÷ 2 = V/2 ÷ 2/1
2
= V/2. x 1/2 = V/4
V/2
Therefore; -------- = V/4
2
So, our V2 ( final velocity, sorry, final volume
) is V/4
V1 = V
V2 = V/4
Now let us start solving.. we have been playing since ooo 
Just kidding, we are almost through..
The percentage change in volume of the gas...
change 100
Precentage change. = ---------- X ------
actual volume 1
Acual volume is same as initial volume..
Change is Initial volume minus final volume
Change = V1 - V2 = V - V/4 = V/1 - V/4
L.C.M is 4.. 4V/4 - V/4 = 3V/4
Therefore change is 3V/4
(You can solve the above solvings your self if u dont understand mine, it is just simple subtraction of fraction)
Recall..
change 100
Precentage change. = ---------- X ------
actual volume 1
Therefore,
3V/4 100
Precentage change. = ---------- X ------
V 1
V cancels V out and the equation becomes...
100
Precentage change. = 3/4 X ------
1
3 X 100
Precentage change. = ----------
4 X 1
300
Precentage change. = ----------
4
Precentage change. = 75 %
Done..
Comment your questions below.
To understand all the things solutions here, just try using each lf the solutions provided. Asin at each step, check which one you understand better and use it.
The answer its simple it's not what u crck ur head for guys!!
now look: and let me cook😋☠️
SOLUTION
let's bring out the parameter first:
V1=V V2=? ( it was not given) T1=T. T2=½T. P1=P. & P2=2P
NOW GIVEN THE FOMULAR AS; P1VI/T1=P2V2/T2
let's substitute parameters into the equation
=P×V/T=2P×V2/0.5T
now make V2 subject of the formular
V2=P×V×0.5/ T×2P
if u cancel out (u know what am talking about, if u don't know better go back to ss1 (cruise ohhh, am not hurting anyone's feeling)🤭
V2=0.5/2
= 0.25V
now remember I the question they said "what is the percentage change in the volume of the gas?"
NOW NOTE THIS THE FORMULAR IS:
increase in volume/original volume ×100
it can be written mathematically as: V1-V2/V1 ×100
now remember
V1=V. V2=0.25.
so substitute parameters into equation
=V-0.25/ V ×100
NOTE: that V in the equation Is also (1)
sooooooo ( let him cook now☠️😹😹😹)
I.e =1-0.25/1
=0.75×100
therefore your answer is= 75%
Please do well to like and comment anywhere u don't get it...😊
not an inkling on this explanation ... please can someone who understands it explain ...
This question is wrong pressure and temperature are directly related/proportional increasing one and reducing the other will not work
