A spherical balloon initially contains 25m3 of helium gas at 20o C and 150 kPa. A valve is now opened and the helium is allowed
to escape slowly. The valve is closed when the pressure inside the balloon drops to the atmospheric pressure of 100 kPa. The elasticity of the balloon material is such that the pressure inside the balloon during the process varies with the volume according to the relation P= A + BV where P is the pressure in kPa, V is the total volume in m3 , the constant A has a value of -100 kPa (i.e. negative 100kPa), and the constant B has an unknown value. The temperature at which the helium exits the balloon is constant at 15o C throughout the whole process.
Disregarding any heat transfer, determine
(a) the final temperature of helium in the balloon after the valve has been closed, and
(b) the mass of helium that has escaped. Consider helium to be an ideal gas with constant specific heats [5.193kJ/(kg K) for Cp,o and 3.116kJ/(kg K) for Cv,o].
Disregarding any heat transfer, determine
(a) the final temperature of helium in the balloon after the valve has been closed, and
(b) the mass of helium that has escaped. Consider helium to be an ideal gas with constant specific heats [5.193kJ/(kg K) for Cp,o and 3.116kJ/(kg K) for Cv,o].
1 answer:
You might be interested in
Hello!
A) At pH=1
This pH is lower than the value for the pKa, so Acetic acid wouldn't be ionized, but the equilibrium would be displaced to CH₃COOH
CH₃COOH ⇄ CH₃COO⁻ + H₃O⁺ (equilibrium displaced to the left)
The chemical structure for CH₃COOH is the first one in the attached images.
B) At pH=7
This pH is higher than the value for the pKa, so Acetic acid would be ionized, and the equilibrium would be displaced to CH₃COO⁻
CH₃COOH ⇄ CH₃COO⁻ + H₃O⁺ (equilibrium displaced to the right)
The chemical structure for CH₃COO⁻ is the second one in the attached images.
Have a nice day!
A) At pH=1
This pH is lower than the value for the pKa, so Acetic acid wouldn't be ionized, but the equilibrium would be displaced to CH₃COOH
CH₃COOH ⇄ CH₃COO⁻ + H₃O⁺ (equilibrium displaced to the left)
The chemical structure for CH₃COOH is the first one in the attached images.
B) At pH=7
This pH is higher than the value for the pKa, so Acetic acid would be ionized, and the equilibrium would be displaced to CH₃COO⁻
CH₃COOH ⇄ CH₃COO⁻ + H₃O⁺ (equilibrium displaced to the right)
The chemical structure for CH₃COO⁻ is the second one in the attached images.
Have a nice day!