A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.
A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.
The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:
The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.
Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.
The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus,
A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.
Learn more: brainly.com/question/22655760
Answer:
(B) II only
Explanation:
In an equilibrium reaction, a keq > 1 means that the reaction is thermodynamically favored. That means the reaction will occurs doing the <em>II affirmation</em> <em>TRUE</em>. But the thermodynamic information doesn't make reference to the rate of the reaction doing <em>I affirmation</em> <em>FALSE</em>. Also, the Keq have relationship with ΔG parameter but ΔH parameter (exothermic or endothermic reaction) is independent of ΔG doing<em> III affirmation</em> <em>FALSE</em>. Thus, right answer is:
<em>(B) II only
</em>
<em />
I hope it helps!
Answer:
I'm pretty sure it's Density
Explanation:
Density differences in water masses due to temperature and salinity variations which makes a process known as thermohaline circulation. These currents move water through the ocean taking nutrients, oxygen, and heat with them.
Answer:
concentration in ppm of DDT = 0.011 ppm
Explanation:
since ppm of DDT =mg of DDT /Kg of solution ( or μg of DDT /g of solution) , then
concentration in ppm of DDT = 4.21 × 10⁻⁷ kg * 10⁶ mg/kg / ( 1.00 g/mL * 3.7 L * 1000 ml/L * 1 kg/1000gr) = 0.011 ppm
then
concentration in ppm of DDT = 0.011 ppm