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Mrrafil [7]
3 years ago
13

A gas is compressed from an initial volume of 5.55 L to a final volume of 1.22 L by an external pressure of 1.00 atm. During the

compression, the gas releases 124 J of heat. What is the change in the internal energy of the gas?
Physics
1 answer:
algol133 years ago
3 0

Answer:

Explanation:

change in the volume of the gas = 5.55 - 1.22

= 4.33 X 10⁻³ m³

external pressure ( constant ) P = 1 x 10⁵ Pa

work done on the gas

=external pressure x change in volume

= 10⁵ x  4.33 X 10⁻³

=4.33 x 10²

433 J

Using the formula

Q = ΔE + W , Q is heat added , ΔE is change in internal energy , W is work done by the gas

Given

Q = - 124 J ( heat is released so negative )

W = - 433 J . ( work done by gas is negative, because it is done on gas  )

- 124  = ΔE - 433

ΔE = 433  - 124

= 309 J

There is increase of 309 J in the internal energy of the gas.

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8 0
3 years ago
Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te
seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy  5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant \gamma =1.4

specific heat is given as =\frac{\gamma R}{\gamma -1}

gas constant =287 J⋅kg−1⋅K−1

Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}

specific heat at constant volume

Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}

change in internal energy = Cv(T_2 -T_1)

                            \Delta U = 717.5 (800-295)  = 3.62*10^5 J kg^{-1}

change in enthalapy \Delta H = Cp(T_2 -T_1)

                                 \Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}

change in entropy

\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})

\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})

\Delta S = 382.79 J kg^{-1} K^{-1}

7 0
2 years ago
An electron in a cathode-ray beam passes between 2.5cm long parallel-plate electrodes that are 6.0mm apart. A 2.1mT, 2.5-cm-wide
Dmitry_Shevchenko [17]

Answer:

(a). The speed of electron is 1.56\times10^{7}\ m/s.

(b). The radius of electron is 4.2\ cm

Explanation:

Given that,

Length = 2.5 cm

Distance = 6.0 mm

Magnetic field = 2.1 T

Potential difference = 700 V

(a). We need to calculate the electron's speed

Using formula of speed

v=\sqrt{\dfrac{2eV}{m}}

Put the value into the formula

v=\sqrt{\dfrac{2\times1.6\times10^{-19}\times700}{9.1\times10^{-31}}}

v=15689290.81\ m/s

v=1.56\times10^{7}\ m/s

(b). We need to calculate the radius of electron

Using formula of centripetal force

\dfrac{mv^2}{r}=qvB

r=\dfrac{mv}{qB}

Where,

m = mass of electron

v = speed of electron

r = radius

q = charge of electron

B = magnetic field

Put the value into the formula

r=\dfrac{9.1\times10^{-31}\times1.56\times10^{7}}{1.6\times10^{-19}\times2.1\times10^{-3}}

r=0.042\ m

r=4.2\ cm

Hence, (a). The speed of electron is 1.56\times10^{7}\ m/s.

(b). The radius of electron is 4.2 cm

8 0
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6 0
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True or false. Atoms of a given element will have the same mass.
jarptica [38.1K]

Answer:

True

Explanation:

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3 years ago
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