THATS SO LONG OMG I WOULD HELP YOU BUT IM NOT GOOD AT READING TOO MUCH
Limit the spread of infectious diseases.
The answer is volt for this question.
I hope is correct if not I’m sorry.
Since the problem is incomplete, I tried to search a similar question in the internet. Luckily, I found the exact problem which is shown in the attached picture. Here are the solution for the five questions:
1. Assuming ideal behavior,
PV = nRT
To use SI units, convert bar to Pa using the conversion: 1 bar = 100,000 Pa. Then, R = 8.314 m³·Pa/mol·K.
(1.15 bar)(100,000 Pa/1 bar)(V) = (0.9 mol)(8.314 m³·Pa/mol·K)(32 + 273 K)
Solving for V,
<em>V = 0.0198 m³</em>2. Since the Pint=Pext because it is allowed to reach equilibrium, P is still 1.15 bar, but T is now 347°C.
PV = nRT
(1.15 bar)(100,000 Pa/1 bar)(V) = (0.9 mol)(8.314 m³·Pa/mol·K)(347 + 273 K)
Solving for V,
<span><em>
V = 0.0403 m³</em></span>
<span>
</span>3. W = PΔV
W = (1.15 bar)(100,000 Pa/1 bar)(
0.0403 m³ - 0.0198 m³)
<em> W = 2,357.5 J or 2.36 kJ</em>5. Let's go first with #5 because this is needed to solve for #4. Internal energy is ΔU.
ΔU = nCvΔT = (0.9 mol)(3.3)(8.314 m³·Pa/mol·K)(347°C - 32°C)
<em>ΔU = 7,778.16 J or 7.78 kJ</em>4. The formula for Q is:
ΔU = Q + W
7.78 kJ = Q + 2.36 kJ
Solving or Q,
<em>Q = 5.42 kJ</em>
1. Average time for the first 0.25 m: 2.23 s
Explanation:
The average time that it takes for the car to travel the first 0.25 m is given by the average of the first three measures, so:
2. Average time to travel between 0.25 m and 0.50 m: 0.90 s
Explanation:
First of all, we need to calculate the time the car takes to travel between 0.25 m and 0.50 m for each trial:
t1 = 3.16 s - 2.24 s = 0.92 s
t2 = 3.08 s - 2.21 s = 0.87 s
t3 = 3.15 s - 2.23 s = 0.92 s
So, the average time is
3. Velocity in the second 0.25 m section: 0.28 m/s
Explanation:
The average velocity in the second 0.25 m section is equal to the ratio between the distance covered (0.25 m) and the average time taken (0.90 s):