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mihalych1998 [28]
4 years ago
6

The decay of a radioactive isotope can be theoretically modeled with the following equation, where C0 is the initial amount of t

he element at time zero and k is the decay rate of the isotope. Create a proper plot of the decay of isotope A [k = 1.48 hours]. Allow time to vary on the abscissa from 0 to 5 hours with an initial concentration of 10 grams of isotope A.

Engineering
1 answer:
liubo4ka [24]4 years ago
6 0

That problem could answered easyly in Matlab,

We know that c=c_0e^{\frac{t}{k}}

So whit this program we can make it,

<em />

<em>figure('color','white')</em>

<em>initial=10;</em>

<em>decay_rate=1.48</em>

<em>time=[0:0:1:5];</em>

<em>concen=initial*exp(-time/decay_rate);</em>

<em>plot(time,concen)</em>

<em>grid</em>

<em>title('Decay of the Isotope A')</em>

<em>xlabel('Time (hours'))</em>

<em>ylabel('concentration in grams')</em>

<em />

<em />

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A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is 20.3 kJ/K
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Answer:

W= 8120 KJ

Explanation:

Given that

Process is isothermal ,it means that temperature of the gas will remain constant.

T₁=T₂ = 400 K

The change in the entropy given ΔS = 20.3 KJ/K

Lets take heat transfer is Q ,then entropy change can be written as

\Delta S=\dfrac{Q}{T}

Now by putting the values

20.3=\dfrac{Q}{400}

Q= 20.3 x 400 KJ

Q= 8120 KJ

The heat transfer ,Q= 8120 KJ

From first law of thermodynamics

Q = ΔU + W

ΔU =Change in the internal energy ,W=Work

Q=Heat transfer

For ideal gas ΔU  = m Cv ΔT]

At constant temperature process ,ΔT= 0

That is why ΔU  = 0

Q = ΔU + W

Q = 0+ W

Q=W= 8120 KJ

Work ,W= 8120 KJ

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