Test: Behaviour of Ideal Gases - JEE MCQ

According to kinetic theory of gases there is no intermolecular force of attraction between the molecule of gases but its not true as gases liquifise, at high pressure molecular interaction starts due to which pressure on the walls decreases as molecule dragged back so pressure exerted by real gas is less than ideal gas, due to repulsive forces molecule behave as small spheres and hence volume decreases

In case of ideal gases the average velocity is always zero. Hence the average momentum is zero. 
Whereas average speed is non- zero so the kinetic energy is also non-zero,  as these two are scalar quantities.

When cp & cv are given calorie & R in Joule then C_p – C_v = (R/J)

1 mole = 6.02 x 10²³

For same mass the ratio of moles of oxygen to that of nitrogen is 14:16 = 7:8
And we know that PV = nRT
Hence as V and P are also same, ratio of temperature of oxygen to that of nitrogen is inverse of the ratio of moles that is 8:7

The pressure law states that for a constant volume of gas in a sealed container the temperature of the gas is directly proportional to its pressure. This can be easily understood by visualising the particles of gas in the container moving with a greater energy when the temperature is increased.

The value of Cp is 30.3
and as Cp-Cv = R(8.3)
hence Cv = 30.3-8.3
Cv is 22
change in internal energy = no of moles × Cv × change in temperature
hence
change in internal energy = 22 × 4 × 10
= 880j
Hence Option D is correct.
 

PV = nRT
Where n = number of moles = m/N_A
So, P = (m/V)(R/N_A)T
Also, we know that R/N_A = k
So, P = nkT

The work done by the gas in taking it from state A to state B = PΔV where ΔW is the increase in volume at constant pressure P. 

We have PV = μRT where p is the number of moles in the sample of the gas and R is the universal gas constant. 

Therefore we have PΔV = μR ΔT = 3 xR(450 - 250) = 600R