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:
Answers A and D are the correct solution.
Explanation:
Both pots have the same mass and the same velocity vector.
the only difference between A and D is the selection of the reference frame positive direction.
the answer is ( True ) .
the current is the same in series circuits .
- Weight (W) = 110 N
- Acceleration due to gravity (g) = 9.8 m/s^2
- Let the mass of the object be m.
- By using the formula, W = mg, we get,
- 110 N = 9.8 m/s^2 × m
- or, m = 110 N ÷ 9.8 m/s^2
- or, m = 11.2 Kg
<u>Answer:</u>
<em><u>The </u></em><em><u>mass </u></em><em><u>of </u></em><em><u>the </u></em><em><u>object </u></em><em><u>is </u></em><em><u>1</u></em><em><u>1</u></em><em><u>.</u></em><em><u>2</u></em><em><u> </u></em><em><u>Kg.</u></em>
Hope you could get an idea from here.
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