What is Ka for H3PO4(aq) ⇌ H+(aq) + H2PO4-(aq) a. Ka = (H3PO4)/(h+)(H2PO4-) B. Ka = (H3PO4)(H+)(H2PO4-) C. Ka = (H3PO4)/(H+)(H2P
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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.
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