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3 years ago

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The volume of a gas is directly related to the temperature of the gas. at 30 degrees celsius a certain gas has a volume of 125 c

ubic centimeters. what will the new volume be at 50 degrees celsius?

2 answers:

pentagon [3]3 years ago

8 0

<u>Answer:</u> The new volume is $133.25cm^3$

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=125cm^3\\T_1=30^oC=(30+273)K=303K\\V_2=?cm^3\\T_2=50^oC=(50+273)K=323K$

Putting values in above equation, we get:

$\frac{125cm^3}{303K}=\frac{V_2}{323K}\\\\V_2=\frac{125\times 323}{303}=133.25cm^3$

Hence, the new volume is $133.25cm^3$

tigry1 [53]3 years ago

5 0

To solve this we use <span>Charles' law whitch states that V1/T1=V2/T2.
So if T1=30 C, V1=123 cm^3 and T2=50 C -> V2=(V1*T2)/T1
T2=208.33 cm^3

</span>

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Answer:

So the specific heat of the liquid B is greater than that of A.

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Liquid A is hotter than the liquid B after both the liquids are heated identically for the same duration of time from the same initial temperature then according to heat equation,

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A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

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Explanation:

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When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

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$\Delta K=\frac{1}{2}mv^2$

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$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

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This force acts as centripetal force, so we can write:

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C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

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From eq.(1), we can rewrite the velocity of the particle as

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$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

D)

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E)

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$r=\frac{mv}{qB}$ (4)

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