Test: The Gas Laws (Old NCERT) - JEE MCQ

Slope of the plot between P and PV at constant temperature is zero. A plot of P v/s PV at constant temperature for a fixed mass of gas is a straight line parallel to the pressure axis. 

The correct answer is option A.
Boyle's law:
In 1662 Robert Boyle studied the relationship between volume and pressure of a gas of fixed amount at constant temperature. He observed that volume of a given mass of a gas is inversely proportional to its pressure at a constant temperature.

The correct answer is option B
Dalton's law (also called Dalton's law of partial pressures) states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases. This empirical law was observed by John Dalton in 1801 and published in 1802.
 

Gay Lussac’s law states that the pressure of a fixed mass of a gas at constant volume is directly proportional to the temperature on Kelvin scale.

The vapour pressure of water will remain the same as the temperature is unchanged. So, the answer is c) 0.4 atm

We have

R = PV/T

= 8.314JK⁻¹mol⁻¹

= 8.314x10⁷ergK⁻¹mol⁻¹

= 2calK⁻¹1mol⁻¹

= 0.0821litreatmK⁻¹mol⁻¹

Hence R depends on units of volume and pressure.

Avogadro's law states that, "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules."

The relationship between the temperature and volume of a gas, which is known as Charles' law, provides an explanation of how hot-air balloons work.

We know, Partial pressure of a gas = mole fraction of that gas x Total pressure of mixture of the gases.

Using this formula and using the information given in this question, we get 
ratio of partial pressures of methane to nitrogen in the mixture as: 1:1

From Charles law we conclude that 

For Boyle’s law, temperature is constant. So the graph is called as isotherm.

N₂O₄ ⇔ 2NO₂
1mole 0 Initial moles
-20% of 1=0.2  +2(20% of 1)=0.40  At equilibrium
1-0.2= 0.8  0.4  Moles at equilibrium
Total moles at equilibrium = 0.8+0.4 = 1.2 moles
1 mole vapour pressure = 1 atm at 300 K
Applying PV = nRT
 1xV = 1x R x 300 .... (1)
When n=1.2 moles, T = 600 K
 P x V = 1.2 x R x 600 .... (2)
Dividing (2) by (1),
PxV/(1xV) = (1.2 x R x 600)/ (1 x R x 300)
Therefore, P= 2.4 atm
Hence, resultant pressure of mixture is 2.4 atm

The correct answer is Option C. 
For ideal gas
pV=RT
V=RT/p
Density is given by
D=M/V
Where M is mass and V is volume
Putting the value of V

P₁V₁=P₂V₂
2=P₂×4
P₂=0.5