Answer:
Step 1;
q = w = -0.52571 kJ, ΔS = 0.876 J/K
Step 2
q = 0, w = ΔU = -7.5 kJ, ΔH = -5.00574 kJ
Explanation:
The given parameters are;
= 100 N·m
= 327 K
= 90 N·m
Step 1
For isothermal expansion, we have;
ΔU = ΔH = 0
w = n·R·T·ln(/) = 1 × 8.314 × 600.15 × ln(90/100) = -525.71
w ≈<em> -0.52571</em> kJ
At state 1, q = w = -0.52571 kJ
ΔS = -n·R·ln(/) = -1 × 8.314 × ln(90/100) ≈ 0.876
ΔS ≈ 0.876 J/K
Step 2
q = 0 for adiabatic process
ΔU = 25×(27 - 327) = -7,500
w = ΔU = <em>-7.5 kJ</em>
ΔH = ΔU + n·R·ΔT
ΔH = -7,500 + 8.3142 × 300 = -5,005.74
ΔH = ΔU = <em>-5.00574 kJ</em>
Answer:
Tone is the author's attitude towards the topic. We can identify it by looking at the setting, characters, details, and word choices. By doing so, it will help us find meaning in the story or passage and help us feel more connected to the writing.
Answer:
the the that keeps the water cycle going is heat
Answer:
Well ads I remember, the motion of the gas particles is random and in a straight-line. A sample of gas is contained in a closed rigid cylinder.
And here is what I found too -
According to Kinetic Molecular Theory, gaseous particles are in a state of constant random motion; individual particles move at different speeds, constantly colliding and changing directions. We use velocity to describe the movement of gas particles, thereby taking into account both speed and direction.
Answer:
Q = -22.9 kJ
Explanation:
Given that,
Mass of water, m = 150.3 g
Water gets cool from 25.60°C to -10.70°C.
The specific heat of water, c = 4.2 J/g°C
The formula for heat needed is given by :
So, 22.9 kJ of heat is needed to be removed to cool.
