0.235 moles of a diatomic gas expands doing 205 j of work while the temperature drops 88 k. find q.? (unit=j)
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Ответ:
The heat exchange in the given process = -806.79 J
Explanation:
0.235 moles of a diatomic gas expands doing 205 J of work while the temperature drops 88 K.
Number of moles , n = 0.235
Temperature difference = 88 K
According to conservation of energy,
ΔQ = ΔU + ΔW
We are given the amount of work done and we have to find the heat exchange in the process.
ΔU =
ΔT
ΔU =
= -601.79 J
Work done by the gas = -205 J
ΔQ = ΔU + ΔW
Substituting the values,
ΔQ = -601.79 + -205 = -806.79 J
The heat exchange in the given process = -806.79 J
Ответ:
6.25 m/s
Explanation:
Given parameters
Initial mark = 65
Final mark = 15
time taken = 8s
Unknown:
Velocity of the runner = ?
Solution:
Velocity is the displacement divided by time;
Velocity =
Displacement = Initial mark - final mark = 65 - 15 = 50m
Now, insert the parameters and solve;
Velocity =
= 6.25m/s