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Softa [21]
3 years ago
10

Calculate ideal work (in J) when a single stream of 1 mole of air is heated and expanded from 25 C and 1 bar to 100 C and 0.5 ba

r. The temperature of the surrounding is 37 C.
Note: Assume air to behave as an ideal gas under these conditions and ideal gas heat capacity is dependent on temperature]
Physics
1 answer:
NISA [10]3 years ago
7 0

Answer:

-1786.5J

Explanation:

Temperature 1=T1=25°c

Temperature 2=T2=200°c

Pressure P1=1bar

Pressure P2=0.5bars

T=37°c+273=310k

Note number if moles=1

Recall work done =2.3026RTlogp2/P1

2.3026*8.314*310log(0.5/1)

-1786.5J

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Explanation:

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3 0
3 years ago
A piece of tape is pulled from a spool and lowered toward a 120-mg scrap of paper. Only when the tape comes within 8.0 mm is the
Anton [14]

Answer:

1.176\times 10^{-3} N

Upward

Explanation:

We are given that

Mass of scarp paper,m=120 mg=120\times 10^{-6}kg

1mg=10^{-6} mkg

Distance,d =8 mm=8\times 10^{-3} m

Magnitude of electric force =F_E= w=mg

Where g=9.8 m/s^2

Substitute the values

F_E=120\times 10^{-6}\times 9.8

F_E=1.176\times 10^{-3} N

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The electric force acts in opposite direction and magnitude of electric force is equal to gravitational force.

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3 years ago
A farmer mated two ducks. They produced three offspring. Two of the offspring were then selected to mate with each other because
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it's called inbreeding


7 0
3 years ago
Capacitor C1 is initially charged to V1 and capacitor C2 is initially charged to V2. The capacitors are then connected to each o
o-na [289]

Answer:

<em>20.08 Volts</em>

Explanation:

<u>Parallel Connection of Capacitors</u>

The voltage across any two elements connected in parallel is the same. If the elements are capacitors, then each voltage is

\displaystyle V_1=\frac{Q_1}{C_1}

\displaystyle V_2=\frac{Q_2}{C_2}

They are both the same after connecting them, thus

\displaystyle \frac{Q_2}{C_2}=\frac{Q_1}{C_1}

Or, equivalently

\displaystyle Q_2=\frac{C_2Q_1}{C_1}

The total charge of both capacitors is

\displaystyle Q_t=Q_1\left(1+\frac{C_2}{C_1}\right)

We can compute the total charge by using the initial conditions where both capacitors were disconnected:

Q_t=V_{10}C_1+V_{20}C_2=25\cdot 24+13\cdot 11=743\ \mu C

Now we compute Q1 from the equation above

\displaystyle Q_1=\frac{Q_t}{\left(1+\frac{C_2}{C_1}\right)}=\frac{743}{\left(1+\frac{13}{24}\right)}=481.95\ \mu C

The final voltage of any of the capacitors is

\displaystyle V_1=V_2=\frac{481.95}{24}=20.08\ V

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