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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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7 0
3 years ago
In a tape recorder, the tape is pulled past the read-and-write heads at a constant speed by the drive mechanism. Consider the re
Irina-Kira [14]

Answer:

Torque decreases .

Explanation:

The tape is pulled at constant speed , speed v is constant , so there is

v = ω r where ω is angular speed and r is radius , As radius decreases , angular speed ω increases , So there is angular acceleration .

Let it be α . Let I be moment of inertia of reel .

Reel is in the form of disc

I = 1/2 m r²

I x α = torque

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6 0
3 years ago
If the pressure exerted on a 300.0 mL sample of hydrogen gas at constant temperature is increased from 0.500 kPa to 0.750 kPa, w
uranmaximum [27]

Answer:

200 mL

Explanation:

Given that,

Initial volume, V₁ = 300 mL

Initial pressure, P₁ = 0.5 kPa

Final pressure, P₂ = 0.75 kPa

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V\propto \dfrac{1}{P}\\\\P_1V_1=P_2V_2

V₂ is the final volume

V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{300\times 0.5}{0.75}\\\\V_2=200\ mL

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5 0
3 years ago
6
hodyreva [135]

Answer:....

Explanation:.

6 0
3 years ago
Determine the binding energy of an F-19 nucleus. The F-19 nucleus has a mass of 18.99840325 amu. A proton has a mass of 1.00728
Anvisha [2.4K]

Answer:

Energy = 1.38*10^13 J/mol

Explanation:

Total number of proton in F-19 = 9

Total number of neutron in F-19 = 10

Expected Mass of F-19  

= 9*1.007 + 10*1.008 = 19.152 u

Actual  mass of F-19 = 18.998 u

Energy of one particle of F-19 = 931.5*Δm = 931.5*(19.152-18.998)

= 143.234 MeV

Energy of one mole of F-19 = 143.234*10^6*1.6*10^-19*6.022*10^23  

= 1.38*10^13 J/mol

8 0
3 years ago
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