a fixed mass of a n ideal gas is heated from 50 to 80C at a constant pressure at 1 atm and again at a constant pressure of 3 atm
1 answer:
Answer:
The required energy remains identical in both scenarios since the specific heat capacity (Cp) does not change with varying pressure.
Explanation:
Given;
initial temperature, t₁ = 50 °C
final temperature, t₂ = 80 °C
Temperature change, ΔT = 80 °C - 50 °C = 30 °C
Pressure for scenario one = 1 atm
Pressure for scenario two = 3 atm
The energy needed in both scenarios is expressed as;
Where;
Cp denotes specific heat capacity, which only varies with temperature and remains unaffected by pressure.
Hence, the energy required remains the same for both scenarios since specific heat capacity (Cp) is pressure-independent.
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Answer:
a) τ = i ^ (y - z
) + j ^ (z Fₓ - x
) + k ^ (x
- y Fₓ)
b) τ = (-0.189i ^ -5.6 j ^ + 33.978k ^) N m
c) α = (-7.8 10⁻⁴ i ^ - 2.3 10⁻² j ^ + 1.4 10⁻¹ k ^) rad / s²
Explanation:
a) The torque can be expressed as
τ = r x F
To tackle this equation, using the determinant approach is the most straightforward method
The resulting expression is
τ = i ^ (y - z
) + j ^ (z Fₓ - x
) + k ^ (x
- y Fₓ)
b) Now let's compute
τ = i ^ (0.075 1.4 -0.035 8.4) + j ^ (0.035 2.8 - 4.07 1.4) + k ^ (4.07 8.4 - 0.075 2.8)
τ = i ^ (- 0.189) + j ^ (-5.6) + k ^ (33,978)
τ = (-0.189i ^ -5.6 j ^ + 33.978k ^) N m
c) To find angular acceleration, we use
τ = I α
α = τ / I
The moment of inertia being a scalar means that only the magnitude of each component changes, orientation remains constant.
α = (-0.189i^ -5.6 j^ + 33.978k^) / 241
α = (-7.8 10⁻⁴ i ^ - 2.3 10⁻² j ^ + 1.4 10⁻¹ k ^) rad / s²