1 A⁰ = 10⁻ ¹⁰ m Atomic volume of 1 mole of hydrogen = Avagadros number × volume of hydrogen molecule = 6.023 × 10²³ × π× (10⁻¹⁰ m)³ = 25.2 × 10⁻⁷ m³ Molar volume = 22.4 L = 22.4 × 10⁻³ m³ Molar volume / Atomic volume = 22.4x10⁻³/25.2x10⁷ = 0.89 × 104 ≈ 104 This ratio is large because actual size ofRead more
1 A⁰ = 10⁻ ¹⁰ m
Atomic volume of 1 mole of hydrogen
= Avagadros number × volume of hydrogen molecule
= 6.023 × 10²³ × π× (10⁻¹⁰ m)³
= 25.2 × 10⁻⁷ m³
Molar volume = 22.4 L = 22.4 × 10⁻³ m³
Molar volume / Atomic volume = 22.4×10⁻³/25.2×10⁷ = 0.89 × 104 ≈ 104
This ratio is large because actual size of gas molecule is negligible in
comparison to the inter molecular separation.
How many kg make 1 unified atomic mass unit?
1u = 1.66 × 10⁻²⁷ kg
1u = 1.66 × 10⁻²⁷ kg
See lessOne mole of an ideal gas at STP occupies 22.4 L. What is the ratio of molar volume to atomic volume of a mole of hydrogen? Why is the ratio so large. Take radius of hydrogen molecule to be 1ºA.
1 A⁰ = 10⁻ ¹⁰ m Atomic volume of 1 mole of hydrogen = Avagadros number × volume of hydrogen molecule = 6.023 × 10²³ × π× (10⁻¹⁰ m)³ = 25.2 × 10⁻⁷ m³ Molar volume = 22.4 L = 22.4 × 10⁻³ m³ Molar volume / Atomic volume = 22.4x10⁻³/25.2x10⁷ = 0.89 × 104 ≈ 104 This ratio is large because actual size ofRead more
1 A⁰ = 10⁻ ¹⁰ m
See lessAtomic volume of 1 mole of hydrogen
= Avagadros number × volume of hydrogen molecule
= 6.023 × 10²³ × π× (10⁻¹⁰ m)³
= 25.2 × 10⁻⁷ m³
Molar volume = 22.4 L = 22.4 × 10⁻³ m³
Molar volume / Atomic volume = 22.4×10⁻³/25.2×10⁷ = 0.89 × 104 ≈ 104
This ratio is large because actual size of gas molecule is negligible in
comparison to the inter molecular separation.
The escape velocity v of a body depends on- (i) the acceleration due to gravity ‘g’ of the planet, (ii) the radius R of the planet. Establish dimensionally the relation for the escape velocity.
v a gᵅ Rᵇ ⇒ v = k gᵅ Rᵇ, K → dimensionless proportionality constant [V] = [g]ᵅ [R]ᵇ [M⁰L¹T⁻¹] = [M⁰L¹T⁻²] [M⁰L¹T⁰]ᵇ equating powers 1 = a + b – 1 = – 2a ⇒ a= 1/2 b= 1 - a = 1-1/2 = 1/2 v=k √gR
v a gᵅ Rᵇ ⇒ v = k gᵅ Rᵇ, K → dimensionless proportionality constant
See less[V] = [g]ᵅ [R]ᵇ
[M⁰L¹T⁻¹] = [M⁰L¹T⁻²] [M⁰L¹T⁰]ᵇ
equating powers
1 = a + b
– 1 = – 2a ⇒ a= 1/2
b= 1 – a = 1-1/2 = 1/2
v=k √gR
A calorie is a unit of heat or energy and it equals 4.2 J where 1J = 1 kg m²s⁻². Suppose we employ a system of units in which unit of mass is α kg, unit of length is β m, unit of time γs. What will be magnitude of calorie in terms of this new system.
n₂ = n₁ [M₁/ M₂]ᵅ (L₁/ L₂)ᵇ (T₁/ T₂)ᶜ = 4.2 ( kg/αkg)¹ (m/βm)² (s/γs)⁻² n₂ = 4.2 α^(-1) β⁻² γ⁻²
n₂ = n₁ [M₁/ M₂]ᵅ (L₁/ L₂)ᵇ (T₁/ T₂)ᶜ
See less= 4.2 ( kg/αkg)¹ (m/βm)² (s/γs)⁻²
n₂ = 4.2 α^(-1) β⁻² γ⁻²
A laser light beamed at the moon takes 2.56s and to return after reflection at the moon’s surface. What will be the radius of lunar orbit
t = time taken by laser beam to go to the moon = distance between earth and moon = d = c × t/2 = 3x10⁸x 2.56/2 = 3.84 x 10⁸ m.
t = time taken by laser beam to go to the moon =
See lessdistance between earth and moon
= d = c × t/2
= 3×10⁸x 2.56/2
= 3.84 x 10⁸ m.