Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure pi and temperature T1 are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to T2. The final pressure pf is:
The intermolecular interaction that is dependent on the inverse cube of the distance between the molecule is:
ion-ion interaction
ion-dipole interaction
London force
London force
B.
ion-dipole interaction
ion-ion interaction is dependent on the square of the distance,
i.e, ion-ion interaction ∝ 1/r2
Similarly,
ion-dipole interaction ∝ 1/r3
London forces ∝ 1/r6
and dipole-dipole interactions ∝ 1/r3
superficially it seems as both ions interaction and hydrogen bonding vary with the inverse cube of the distance between the molecules but when we look at the exact expressions of field (force) created in two situations it comes as,
In the case of ion-dipole interaction:
and, In the case of dipole-dipole interaction
From the above it is very clear, the ion-dipole interaction is the better answer as compared dipole-dipole interaction i.e. hydrogen bonding.
If Z is a compressibility factor, Vander Waal's equation at low pressure can be written as
For the gaseous state, if most probable speed is denoted by C*, average speed by C and mean square speed by C, then for a large number of molecules the ratios of these speeds are:
'a' and 'b' are van der Waals constants for gases. Chlorine is more easily liquefied than ethane because :
a and b for Cl2 > a and b for C2H6
a and b for Cl2 < a and b for C2H6
a and Cl2 < a for C2H6 but b for Cl2 > b for C2H6
a and Cl2 < a for C2H6 but b for Cl2 > b for C2H6
If 10–4 dm3 of water is introduced into a 1.0 dm3 flask at 300 K, how many moles of water are in the vapour phase when equilibrium is established?(Given: Vapour pressure of H2O at 300 K is 3170 Pa; R = 8.314 J K–1 mol–1)
5.56 x 10-3 mol
1.53 x 10-2 mol
4.46 x 10-2 mol
4.46 x 10-2 mol
Assuming that water vapour is an ideal gas, the internal energy change(∆U) when 1 mol of water is vapourised at 1 bar pressure and 100°C, (Given: Molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol-1 and R = 8.3 J mol–1K–1 will be) –
4.100 kJ mol–1
3.7904 kJ mol–1
37.904 kJ mol–1
37.904 kJ mol–1
Phosphorus pentachloride dissociates as follows, in a closed reaction vessel,
PCl5 (g) ⇌ PCl3(g) + cl2(g)
If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the partial pressure of PCl3 will be
The formation of the oxide ion O2-(g) requires first an exothermic and then an endothermic step as shown below
Oxygen is more electronegative
O- ion has comparatively larger size than oxygen atom
O- ion will tend to resist the addition of another electron
O- ion will tend to resist the addition of another electron