Answer:
(A) 10132.5Pa
(B)531kJ of energy
Explanation:
This is an isothermal process. Assuming ideal gas behaviour then the relation P1V1 = P2V2 holds.
Given
m = 10kg = 10000g, V1 = 0.1m³, V2 = 1.0m³
P1 = 101325Pa. M = 102.03g/mol
P2 = P1 × V1 /V2 = 101325 × 0.1 / 1 = 10132.5Pa
(B) Energy is transfered by the r134a in the form of thw work done in in expansion
W = nRTIn(V2/V1)
n = m / M = 10000/102.03 = 98.01mols
W = 98.01 × 8.314 × 283 ×ln(1.0/0.1)
= 531kJ.
Answer: C.
Explanation:
i choose B and got it wrong it told me the answer after on edge.
Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)
• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)
•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)
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Answer:
529.15 m/s
Explanation:
h = Maximum height = 70000 m
g = Acceleration due to gravity = 2 m/s²
m = Mass of sulfur
As the potential and kinetic energies are conserved

The speed with which the liquid sulfur left the volcano is 529.15 m/s