Part 1- The work done by the gas during this process will be 5.65 ×10⁻³ kJ.
Part 2-The heat added to the gas during this process will be 5.65 ×10×10⁻³ kJ.
<h3>What is work done by the gas?</h3>
Work is the product of pressure p and volumes V during a volume change for such a gas. The work seems to be the area under the curve that indicates how the state changes.
The work done under the isothermal process is;
For the isothermal process;
ΔU=0
Hence, the work done, and the heat added by the gas during this process will be 5.65 ×10⁻³ kJ and 5.65 ×10×10⁻³ kJ respectively.
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Well, It rather depends on your definition of "machine." The normal physics set of simple machines - levers, pulleys, ramps all give you increased the force at the expense of reduced speed or increased the rate at the cost of reduced force. So, no - by definition a machine is an arrangement for multiplying one while paying the cost by reducing the other. You are looking at an example of the Conservation of Energy. One of the giant rules we are pretty sure cannot be violated.<span>
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To solve this problem, we have to use the formula:
E = h f
where E is total energy, h is Plancks constant
6.626x10^-34 J s, f is frequency
f = E / h
f = 3.686 × 10−24 J / (6.626x10^-34 J s)
<span>f = 5.56 x 10^9 Hz</span>
If you, for example, poured it onto a wide cup with a volume equal to the total volume of the sand particles, the sand would not spread out to fill the container but would bunch up together in the middle.
