Answer:
a)P₂ =4 bar
b)W= - 1482.48 KJ
It means that work done on the system.
c)S₂ - S₁ = 3.42 KJ/K
Explanation:
Given that
T₁ = 300 K ,V₁ = 3 m³ ,P₁=2 bar
T₂ = 600 K ,V₂=V₁ 3 m³
Given that tank is rigid and insulated.It means that volume of the gas will remain constant.
Lets take the final pressure = P₂
For ideal gas P V = m R T
P₂ =4 bar
Internal energy
ΔU = m Cv ΔT
Cv=0.71 KJ/kg.k for air
m= 6.96 kg
ΔU= 6.96 x 0.71 x (600 - 300)
ΔU=1482.48 KJ
From first law
Q= ΔU + W
Q= 0 Insulated
W = - ΔU
W= - 1482.48 KJ
It means that work done on the system.
Change in the entropy
S₂ - S₁ = 3.42 KJ/K
Answer:
d. 50 C
Explanation:
In this problem, we have to add 800 ml of water at 20 Celsius to 800 ml of water at 80 Celsius.
According to the 2nd law of thermodynamics, heat transfers from hot to cold temperature.
The quantity of both the different waters is equal so this makes it very easy. All we have to do is find the mean of both the temperatures:
Final temperature = (20 C + 80 C)/2
= 50 Celsius
Usually start on the internet, there is bound to be something or a form of information on it.
Answer:
a) the magnitude of the force is
F= Q(
) and where k = 1/4πε₀
F = Qqs/4πε₀r³
b) the magnitude of the torque on the dipole
τ = Qqs/4πε₀r²
Explanation:
from coulomb's law
E = 
where k = 1/4πε₀
the expression of the electric field due to dipole at a distance r is
E(r) = 
, where p = q × s
E(r) = 
where r>>s
a) find the magnitude of force due to the dipole
F=QE
F= Q()
where k = 1/4πε₀
F = Qqs/4πε₀r³
b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces
τ = F sinθ × s
θ = 90°
note: sin90° = 1
τ = F × r
recall F = Qqs/4πε₀r³
∴ τ = (Qqs/4πε₀r³) × r
τ = Qqs/4πε₀r²
