Test: Boyle's & Charles Law (Old NCERT) - JEE MCQ

Since the system is isothermal, so temperature is constant.
According to Boyle's lawP₁V₁= P₂V₂
20×0.5=1×V₂
10=V₂

V1 = 10xcm³
V2 = ?
P1 = 1atm = 760mmHg
P2 = (760 + 4)mmHg
P1V1 = P2V2
V2 = P1V1 / P2
= 760 × 10x / 764
= 9.5x
Therefore,
= 9.5x / x = 9.5

Applying Boyle 's law P1V1=P2V2 as temp is constant.
So let initial pressure =P initial volume =V 
Now final volume V2=V-10%of V =9V/10
PV=P2×9V/10 so P2=10P/9
Therefore increment in Pressure=P/9P×100%=100/9

We know ideal gas equation,
PV=nRT.
n= m(mass)/M(molecular weight).
PV=m/M RT.
PM=m/v RT.
density (d)=m(mass)/v (volume).
PM=dRT.
d=PM/RT.
here, density is directly proportional to pressure and inversely proportional to temperature...
d₁/d₂=P₁ T₂ / P₂ T₁.
1.25/d₂= 1×298/1.5×298.
1.25/d₂=1/1.5.
d₂= 1.25×1.5.
d₂=1.875

We know that
pV = nRT 
Or V = nRT/p
If we increase both T and P 2 times, only then the volume remains constant.

Let the number of moles of dihydrogen and dioxygen be 1 and 4.

According to Charles law,at constant pressure, V∝T
Or V/T = k

In this setup
The liquid in tube 1 has got lowered by height h1 which means that the pressure inside the tube is greater than the pressure exerted by the atmosphere. This excess pressure must be equal to the pressure exerted by h1 column of liquid of the same cross section ,therefore
Pressure inside tube 1=atmospheric pressure+h1dg
In tube 2 since the level of water is same as that outside the tube so her
Pressure in tube 2=atmospheric pressure
In tube 3 the liquid has risen by height be which means that pressure inside the tube is less than the pressure exerted by atmospheric pressure .
Atmospheric pressure=pressure in tube+h₃dg
Pressure in tube=atmospheric pressure -h₃dg

From  ideal gas equation, we have
pV = nRT
Or V = nRT/p
Taing log on both sides
log V = log nRT - log p
Or y = mx - c
Let us make a point constant on time axis like this

Now we have the temperature constant.So to have maximum value of log V, we need to have minimum value of log p. From the graph we can see that we have least pressure for p3 . So the order will be p₃<p₂<p₁.

-272deg. 
For C = 1 K
V / 1 = K
V = K

From Boyle’s law;
p ∝ 1/V ⇒ Graph i (Straight line passing through origin)
Or pV = Constant ⇒ Graph ii (rectangular hyperbola)
Taking log on both side;
log(pV) = logK
logp + logV = k’
logp = -logV = k’ (y = -mx+c) ⇒ Graph iii
log p = log V-1 + k’
Logp = log(1/V) + k’ ⇒ Graph iv
So option c is correct

α = -1/V(dV/dP)_T
pV = RT;
V = RT/p
dV/dp = -RT/p²
α = -1/(RT/p)×(-RT/p²)
α = 1/p
Or α =  (since p = 0.2)