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den301095 [7]
3 years ago
5

In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 2820 J of work is done on the gas

.
Required:
a. How much heat flows into or out of the gas?
b. What is the change in internal energy of the gas?
c. Does its temperature rise or fall?
Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
3 0

Answer:

  1. Q=0
  2. U=2820
  3. Energy increases

Explanation:

From the question we are told that

Work done W=2820

a)Generally the heat flow for an adiabatic process is 0 (zero)

Q= U + W =>0

Q=0

b)Generally Change in internal energy of gas is mathematically given by

Since W=-2820J

Therefore

U=2820

Giving

Q= 2820 -2820

Q=O

c)With increases in internal energy brings increase in temperature

Therefore

Energy increases

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E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

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