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IrinaK [193]
3 years ago
11

An ideal gas, consisting of n moles, undergoes an irreversible process in which the temperature has the same value at the beginn

ing and end. If the volume changes from Vi to Vf , the change in entropy is given by:______
Engineering
1 answer:
Vlada [557]3 years ago
4 0

Answer:

n R ln(Vf/Vi)

Explanation:

Entropy is the loss of energy available to do work. Entropy is a state function (i.e. it depends only upon the current state of the system and is independent of how that state was prepared).

Since the temperature change of the ideal is constant, hence this is an isothermal expansion of a perfect gas. The change in entropy (ΔS) for an isothermal expansion of a perfect gas is given by:

\Delta S=nR*ln(\frac{V_f}{V_i})

Where n is the amount of gas molecules in mol and R is the gas constant in JK⁻¹mol⁻¹given by R = N_Ak, k is Boltzmann's constant in J K⁻¹ and Avogadro's constant N_A in mol⁻¹. Vf is the final volume and Vi the initial volume.

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complete question

A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?

Answer:

3.03 V  0.184 W

2.499 mV  125*10^-9 W

Explanation:

First, apply voltage-divider principle to the input circuit: 1

V_{i}= (R_i/R_i+R_s) *V_s = 10^6/10^6+(0.1*10^6)\\*5

    = 4.545 V

The voltage produced by the voltage-controlled source is:

A_voc*V_i = 4.545 V

We can find voltage across the load, again by using voltage-divider principle:  

V_o = A_voc*V_i*(R_o/R_l+R_o)

      = 4.545*(100/100+50)

      = 3.03 V  

Now we can determine delivered power:  

P_L = V_o^2/R_L

      = 0.184 W

Apply voltage-divider principle to the circuit:  

V_o = (R_o/R_o+R_s)*V_s

       = 50/50+100*10^3*5

       = 2.499 mV

Now we can determine delivered power:  

P_l = V_o^2/R_l

     = 125*10^-9 W

Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.  

4 0
3 years ago
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