Answer:
the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)
Explanation:
assuming ideal gas behaviour of the gas , the equation for ideal gas is
P*V=n*R*T
where
P = absolute pressure
V= volume
T= absolute temperature
n= number of moles of gas
R= ideal gas constant = 8.314 J/mol K
P=n*R*T/V
the work that is done by the gas is calculated through
W=∫pdV= ∫ (n*R*T/V) dV
for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:
W=∫pdV= ∫ (n*R*T/V) dV = n*R*T ∫(1/V) dV = n*R*T * ln (V₂/V₁)
since
P₁=n*R*T/V₁
P₂=n*R*T/V₂
dividing both equations
V₂/V₁ = P₁/P₂
W= n*R*T * ln (V₂/V₁) = n*R*T * ln (P₁/P₂ )
replacing values
P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa
since P₂ = 1 atm = 101325 Pa
W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J
Answer:
<em>The velocity of the camera just before it hit the ground = 35.97 m/s.</em>
Explanation:
Velocity: This can be defined as the ratio of the displacement of a body to the time. Velocity is a vector quantity, and as such it can be represented both in magnitude and direction.
From the equation of motion,
v² = u² + 2gs ................ Equation 1
Where v = final velocity, u = initial velocity, g = acceleration due to gravity, s = distance.
Note: Before the velocity of the camera before it hits the ground = The final velocity of the camera.
<em>Given: u = 10.8 m/s, s = 60 m. g = 9.81 m/s.</em>
<em>Substituting into equation 1,</em>
<em>v² = 10.8² + 2(9.81)(60)</em>
<em>v² = 116.64+1177.2</em>
<em>v² = 1293.84 </em>
<em>v = √(1293.84)</em>
<em>v = 35.97 m/s.</em>
<em>Hence the velocity of the camera just before it hit the ground = 35.97 m/s.</em>
Answer:
1.19cm^3 of glycerine
Explanation:
Let Vo= 150cm^3 for both aluminum and glycerine, using expansion formula:
Volume of spill glycerine = change in volume of glycerine - change in volume of aluminum
Volume of glycerine = coefficient of volume expansion of glycerine * Vo* change in temperature - coefficient of volume expansion of Aluminum*Vo* change temperature
coefficient of volume expansion of aluminum = coefficient of linear expansion of aluminum*3 = 23*10^-6 * 3 = 0.69*10^-4 oC^-1
Change in temperature = 41-23 = 18oC
Volume of glycerine that spill = (5.1*10^-4) - (0.69*10^-4) (150*18) = 4.41*10^-4*2700 = 1.19cm3
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Answer:
10
Explanation:
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