The energy becomes 0.50 times in 6.72 s.
Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.
Et=0.50E
1
k(4₂)² = (0.5) - kA²
(4₂)² = (0.5) A²
At = 0.71A
So, the amplitude of the oscillator becomes 0.71 times its initial ar
0.71A = = A(0.96)¹2
log(0.71)
log(0.96)
8.4
n=
So, the time taken for n oscillation is obtained as,
t = n (0.800 s)
= (8.4) (0.800)
= 6.72 s
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Answer:
T2=336K
Explanation:
Clausius-Clapeyron equation is used to determine the vapour pressure at different temperatures:
where:
In(P2/P1) = ΔvapH/R(1/T1 - 1/T2)
p1 and p2 are the vapour pressures at temperatures
T1 and T2
ΔvapH = the enthalpy of vaporization of the liquid
R = the Universal Gas Constant
p1=p1, T1=307K
p2=3.50p1; T2=?
ΔvapH=37.51kJ/mol=37510J/mol
R=8.314J.K^-1moL^-1
In(3.50P1/P1)= (37510J/mol)/(8.314J.K^-1)*(1/307 - 1/T2)
P1 and P1 cancelled out:
In(3.50)=4511.667(T2 - 307/307T2)
1.253=14.696(T2 - 307/T2)
1.253=(14.696T2) - (14.696*307)/T2
1.253T2=14.696T2 - 4511.672
Therefore,
4511.672=14.696T2 - 1.253T2
4511.672=13.443T2
So therefore, T2=4511.672/13.443=335.61
Approximately, T2=336K
Answer:
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Explanation:
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