The pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.
Therefore, option A is correct option.
Given,
Mass m = 14g
Volume= 3.5L
Temperature T= 75+273 = 348 K
Molar mass of CO = 28g/mol
Universal gas constant R= 0.082057L
Number of moles in 14 g of CO is
n= mass/ molar mass
= 14/28
= 0.5 mol
As we know that
PV= nRT
P × 3.5 = 0.5 × 0.082057 × 348
P × 3.5 = 14.277
P = 14.277/3.5
P = 4.0794 atm
P = 4.1 atm.
Thus we concluded that the pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.
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Even though the wind "tries" to flow from high pressure to low pressure, the turning of the Earth causes the air flow to turn to the right (in the Northern Hemisphere), so the jet stream flows around the air masses, rather than directly from one to the other.
Answer:
The force on one side of the plate is 3093529.3 N.
Explanation:
Given that,
Side of square plate = 9 m
Angle = 60°
Water weight density = 9800 N/m³
Length of small strip is
The area of strip is
We need to calculate the force on one side of the plate
Using formula of pressure
On integrating
Hence, The force on one side of the plate is 3093529.3 N.
Answer:
Explanation:
Force between two charges of q₁ and q₂ at distance d is given by the expression
F = k q₁ q₂ / d₂
Here force between charge q₁ = - 15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = (1.66 - 1.24 ) = .42 mm
k = 1/ 4π x 8.85 x 10⁻¹²
putting the values in the expression
F = 1/ 4π x 8.85 x 10⁻¹² x - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²
= 9 x 10⁹ x - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²
= 35969.4 x 10⁻³ N .
force between charge q₂ = 34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = ( 1.24 - 0 ) = 1.24 mm .
putting the values in the expression
F = 1/ 4π x 8.85 x 10⁻¹² x 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²
= 9 x 10⁹ x - 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²
= 82729.6 x 10⁻³ N
Both these forces will act in the same direction towards the left (away from the origin towards - ve x axis)
Total force = 118699 x 10⁻³
= 118.7 N.
Answer:
Explanation:
We know,
..............(1)
where,
η = Efficiency of the engine
T₁ = Initial Temperature
T₂ = Final Temperature
Q₁ = Heat available initially
Q₂ = Heat after reaching the temperature T₂
Given:
η =0.280
T₁ = 3.50×10² °C = 350°C = 350+273 = 623K
Q₁ = 3.78 × 10³ J
Substituting the values in the equation (1) we get
or
or
⇒ 
Now,
The entropy change (
) is given as:
or
substituting the values in the above equation we get
