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:
1201 lbs
Explanation:
Given that in mammals, the weight of the heart is approximately 0.5% of the total body weight.
Let the weight of the heart of a mammal be H
And the weight of the total body be B
The linear model that can gives the heart weight in terms of the total body weight will be:
H = 0.005B
B.) To find the weight of the heart of a whale whose weight is 2.402 × 105 lbs, substitute the whole weight in the formula.
H = 0.005 × 2.402 × 10^5
H = 1201 lbs
Therefore, the weight of the heart of the whale is 1201 lbs
Answer:
v = 8.57 m/s
Explanation:
As we know that the wagon is pulled up by string system
So the net force on the wagon along the inclined is due to tension in the rope and component of weight along the inclined plane
So as per work energy theorem we know that
work done by tension force + work done by force of gravity = change in kinetic energy
so we have
m = 38.2 kg
d = 85.4 m
so now we have