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nadya68 [22]
3 years ago
7

2. A compression at a constant pressure of 20 kPa is performed on 9.0 moles of an ideal monatomic gas (Cv-1.5R). The compression

reduces the volume
of the gas from 0.26 m3 to 0.12 m3. What is the change in the internal energy of the gas? Let the ideal-gas constant R = 8.314 J/(mol. K).
0 -7.0 kj
O zero
4.2 kJ
O-4.2 k]
O 7.0 kj
Physics
1 answer:
Fofino [41]3 years ago
8 0

Answer:

- 4.2 kJ

Explanation:

We shall find out the initial temperature of the gas

PV / T = n R

20 X 10³ X .26 / T = 9 X 8.3

T = 69.6 K

Since the pressure of gas is constant

V / T = constant

V₁ / T₁ = V₂ / T₂

.26 / 69.6 = .12 / T₂

T₂ = 32.12 K

gas is cooling so there will be decrease in internal energy

Change in temperature = 69.6 - 32.12

= 37.48 K.

decrease in internal energy

= - n Cv x fall in temperature

= - 9 x 1.5 x 8.31 x 37.48  ( Cv = 1.5 R given )

= - 4.2 kJ .

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Explanation:

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3 years ago
A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi
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0.099C

Explanation:

First, we need to get the common potential voltage using the formula

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C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}

Therefore

V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437

Charge, Q is given by CV hence for the first capacitor charge will be Q_1=C_1V

Here, Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C

8 0
3 years ago
The burning of a log releases the logs chemical_energy into other forms of energy
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Please answer this question for me and explain why.
horsena [70]

Answer:

D.None of these

Explanation:

The derivation of acceleration formula:

Let us call the 5kg mass m_2 and the 4kg mass m_1. If the tension in the string is T then for the mass m_2

(1). T-m_2g=-m_2a <em>(the negative sign on the right side indicates that acceleration is downwards)</em>

And for the mass m_1

(2). T-m_1g =m_1a<em> (the acceleration is upwards, hence the positive sign)</em>

Solving for T in the 2nd equation we get:

T =m_1a+m_1g,

and putting this into the 1st equation we get:

m_1a+m_1g-m_2g=-m_2a\\\\m_1a+m_2a = m_2g-m_1g\\\\a(m_1+m_2)= (m_2-m_1)g

\boxed{a= \dfrac{(m_2-m_1)}{(m_1+m_2)} g}

Back to the question:

Using the formula for the acceleration we find

a= \dfrac{(5kg-4kg)}{(5kg+4kg)} g

a = \dfrac{g}{9},

which is the acceleration that none of the given choices offer. Also, the acceleration of the two blocks is the same, because if it weren't, the difference in the instantaneous velocities of the objects would cause the string to break. Therefore, these two reasons make us decide that none of the choices are correct.

7 0
3 years ago
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