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Nat2105 [25]
3 years ago
14

A fixed amount of ideal gas is kept in a container of fixed volume. The absolute pressure P, in pascals, of the gas is plotted a

s a function of its temperature T, in degrees Celsius. Which of the following are properties of a best fit curve to the data? Select two answers.
A) Having a positive slope
B) Passing through the origin
C) Having zero pressure at a certain negative temperature
D) Approaching zero pressure as temperature approaches infinity
Physics
1 answer:
Vesnalui [34]3 years ago
6 0

Answer:

A) Having a positive slope

D) Approaching zero pressure as temperature approaches infinity

Explanation:

A fixed amount of ideal gas is kept in a container of fixed volume. The absolute pressure P, in pascals, of the gas is plotted as a function of its temperature T, in degrees Celsius. Which of the following are properties of a best fit curve to the data? Select two answers.

A) Having a positive slope

B) Passing through the origin

C) Having zero pressure at a certain negative temperature

D) Approaching zero pressure as temperature approaches infinity

solution

from the ideal gas equation

PV=nRT

P=pressure

T=absolute temperature in kelvin

V=volume

R=gas constant 8.314j/molK

n=mole= mass/molecular mass

P=nRT/V

nR/V is the constant

P=nR(Tc+273)/V

plotting the graph we will discover that as pressure increases , temperature also increases . which means it's a positive graph.

Also  Approaching zero pressure as temperature approaches infinity

You might be interested in
5.54 Two kilograms of air within a piston–cylinder assembly WP execute a Carnot power cycle with maximum and minimum temper atur
Step2247 [10]

Answer:

A) 60%

B) p2 = 1237.2 kPa

   v2 = 0.348 m^3

C) w1-2 = w3-4 = 1615.5 kJ

   Q2-3 = 60 kJ

Explanation:

A) calculate thermal efficiency

  Л = 1 - \frac{Tl}{Th}  

where Tl = 300 k

            Th = 750 k

hence thermal efficiency ( Л ) = [1 - ( 300 / 750 )] * 100 = 60%

B) calculate the pressure and volume at the beginning of the isothermal expansion

calculate pressure ( P2 )  :

= P3v3 = mRT3  ----- (1)

v3 = 0.4m , mR = 2* 0.287, T3 = 750

hence P3 = 1076.25

next equation to determine P2

Qex = p3v3 ln( p2/p3 )

60 = 1076.25 * 0.4 ln(p2/p3)

hence ; P2 = 1237.2 kpa

calculate volume ( V2 )

p2v2 = p3v3

v2 = p3v3 / p2

   = (1076.25 * 0.4 ) /  1237.2  

  = 0.348 m^3

C) calculate the work and heat transfer for each four processes

work :

W1-2 = mCv( T2 - T1 )

        = 2*0.718 ( 750 - 300 ) = 1615.5 kJ

W3-4 = 1615.5 kJ

heat transfer

Q2-3 = W2-3 = 60KJ

Q3-4 = 0

D ) sketch of the cycle on p-V coordinates

attached below  

6 0
3 years ago
A particle with charge 6 mC moving in a region where only electric forces act on it has a kinetic energy of 1.9000000000000001 J
Vesna [10]

Answer:

16.9000000000000001 J

Explanation:

From the given information:

Let the initial kinetic energy from point A be K_A = 1.9000000000000001 J

and the final kinetic energy from point B be K_B = ???

The charge particle Q = 6 mC = 6 × 10⁻³ C

The change in the electric potential from point B to A;

i.e. V_B - V_A = -2.5 × 10³ V

According to the work-energy theorem:

-Q × ΔV = ΔK

-Q \times ( V_B - V_A) = (K_B - K_A)

-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)

15 = (K_B - 1.9000000000000001 \ J)

K_B = 15+ 1.9000000000000001 \ J

\mathbf{K_B =1 6.9000000000000001 \ J}

3 0
3 years ago
Find the kinetic energy of a 0.1 kg toy truck moving at a speed of 1.1 m/s
enot [183]
0.0605J is your answer. Use the formula KE=1/2mv^2
8 0
4 years ago
Identify the palindromic base sequence in the DNA donor molecule shown.
Airida [17]

Explanation:

Since a double helix is formed by two paired antiparallel strands of nucleotides that run in opposite directions, and the nucleotides always pair in the same way (adenine (A) with thymine (T) in DNA or uracil (U) in RNA; cytosine (C) with guanine (G)), a (single-stranded) nucleotide sequence is said to be a palindrome

6 0
3 years ago
A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision
marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

x \approx 103,46x10^{9}m.

B) The space time coordinate t of the collision in Earth's reference frame is

t=377,29s

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

                          x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                                y'=y

                                z'=z

                          t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

                       x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                           y=y'

                           z=z'

                        t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

First we calculate the expression in the denominator

                            \frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}

                                \sqrt{1-\frac{v^{2}}{c^{2}}} =0,714

then we calculate t

                      t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                      t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}

                      t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}

                      t=\frac{190s+79,388s}{0,714}

finally we get that

                                     t=377,29s

then we calculate x

                         x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                         x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+39872396914m}{0,714}}

                         x=\frac{73872396914m}{0,714}}

                         x=103462740775,91m

finally we get that

                                     x \approx 103,46x10^{9} m

5 0
4 years ago
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