The beans are cooked earlier in pressure cooker, because

boiling point increases with increasing pressure
boiling point decreases with increasing pressure
internal energy is not lost while cooking in pressure cooker
extra pressure of pressure cooker, softens the beans

Solution

The beans are cooked earlier in pressure cooker because boiling point increases with increasing pressure.

The term that corrects for the attractive forces present in areal gas in the van der Waals equation is

nb
^an²⁄_V²
-^an²⁄_V²
-nb

Solution

Correction factor for attractive force for n moles of real gas is given by the term mentioned in (b).

For the gas A, a = 0 and its dependence on P is linear at all pressure.
For the gas B, b = 0 and its dependence on P is linear at all pressure
For the gas C, which is typical real gas for which neither a nor b = 0. By knowing the minima and the point of intersection, with Z = 1, a and b can be calculated
At high pressure, the slope is positive for all real gases

The compression factor (compressibility factor) for 1 mole of a van der Waal’s gas at 0°C and 100 atm pressure if found to be 0.5. Assuming that the volume of gas molecules is negligible, calculate the van der Waal’s constant ‘a’.

0.253 L²mol^-2atm
0.53 L²mol^-2atm
1.853 L²mol^-2atm
1.253 L²mol^-2atm

Solution

Solution

The compressibility factor for areal gas at high pressure is:

\(1 + \frac{RT}{pb}\)
1
\(1 + \frac{pb}{RT}\)
\(1 - \frac{pb}{RT}\)

Solution

When r, P and M represent rate of diffusion, pressure and molecular mass, respectively, then the ratio of the rates of diffusion (r_A/r_B) of two gases A and B, is given as :

(P_A/P_B)(M_B/M_A)^1/2
(P_A/P_B)^1/2(M_B/M_A)
(P_A/P_B)(M_A/M_B)^1/2
(P_A/P_B)^1/2(M_A/M_B)

Solution

50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar condition. If molecular mass of gas B is 36, the molecular mass of gas A will be :

96
128
20.25
64

Solution

A bubble of air is underwater at temperature 15°C and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is 25°C and the pressure is 1.0 bar, what will happen to the volume of the bubble?

Volume will be come greater by a factor of 1.6.
Volume will become greater by a factor of 1.1.
Volume will become smaller by a factor of 0.70.
Volume will become greater by a factor of 2.5.

Solution

Given

P₁ = 1.5 bar T₁ = 273 + 15 = 288K V₁ = V

P₂ = 1.0 bar T₂ = 273 + 25 = 298K V₂ = ?

\(\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\)

\(\frac{1.5 \times V}{288} = \frac{1 \times V_{2}}{298}\)

V₂ = 1.55 V i.e., volume of bubble will be almost 1.6 time to initial volume of bubble.

By what factor does the average velocity of a gaseous molecule increase when the temperature (in Kelvin) is doubled?

Solution

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