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earnstyle [38]
3 years ago
8

Container A and container B hold samples of the same ideal gas. The volume and the pressure of container A is equal to the volum

e and pressure of container B, respectively. If Container A has half as many molecules of the ideal gas in it as Container B does, then which of the following mathematical statements is correct regarding the absolute temperatures TA and TB in Container A and Container B. respectively?
A. TA = TB/2.
B. TA = 4TB.
C. TA = TB/4.
D. TA = 2TB.
E. TA = TB
Physics
1 answer:
SVETLANKA909090 [29]3 years ago
6 0

Answer:

A. TA = TB/2.

Explanation:

Since container A has half as many molecules of the ideal gas in it as container B. Therefore, container A will have half the volume of gas as in container B:

V_A = \frac{1}{2}V_B

Now, from Charle's Law:

\frac{V_A}{T_A}=\frac{V_B}{T_B}\\\\\frac{1}{2}\frac{V_B}{T_A}=\frac{V_B}{T_B}\\\\T_A = \frac{T_B}{2}

Hence, the correct option is:

<u>A. TA = TB/2.</u>

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Volume of given mass of gas is inversely proportional to pressure of gas

Explanation:

Boyle's law: It states that the volume of given mass of gas is inversely proportional  to the pressure of gas at constant temperature.

Mathematical representation:

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8 0
4 years ago
A student is experimenting with some insulated copper wire and a power supply. She winds a single layer of the wire on a tube wi
OverLord2011 [107]

Answer:

P=214.7187\,W

Explanation:

Given that:

Diameter of the solenoid, D=10\,cm=0.1\,m

length of the solenoid, L=90\,cm=0.9\,m

diameter of the wire, d=0.1\,cm=10^{-3}\,m

magnetic field at the center of the solenoid, B=7.4\times 10^{-3}\,T

<u>Now we need the no. of turns incorporated in the length of 90 cm:</u>

N=\frac{Length\,\,of\,\,solenoid}{diameter\,\,of\,\, wire}

N=\frac{L}{d}

N=\frac{0.9}{10^{-3}}

N=900\,\,turns

For solenoids we have:

B=\mu.n.I ...............................(1)

where:

\mu=permeability of the medium

n = no. of turns per unit length

I = current in the coil

So,

n=\frac{900}{0.9}

n=1000\,turns\,.\,m^{-1}

Now putting the respective values in the eq. (1)

7.4\times 10^{-3}=4\pi\times10^{-7}\times 1000\times I

I=5.8887\,A

  • For copper we have resistivity:
  • \rho=1.72\times 10^{-8}\, \Omega.m

We know that resistance is given by:

R=\rho.\frac{l}{a} .....................................(2)

where:

l = length of the conducting wire

a = cross sectional area of the conducting wire

<u>Now we need the length (l) of the wire:</u>

Circumference of the solenoid,

C=\pi.D

C=0.1\pi\,m

\therefore l=C\times N

l=90\pi\,m

&

<u>Cross-sectional area of wire:</u>

a=\pi.\frac{d^2}{4}

a=\pi. \frac{(10^{-3})^2}{4}\,m^2

<u>Resistance from eq. (2):</u>

R=1.72\times 10^{-8}\times \frac{90\pi}{\pi. \frac{(10^{-3})^2}{4}}

R=6.192 \,\Omega

  • For power we have:

P=I^2.R

P=5.8887^2 \times 6.192

P=214.7187\,W

6 0
3 years ago
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