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eimsori [14]
2 years ago
9

The volume of an ideal gas is decreased from 5.0 l to 5.0 ml at a constant pressure of 2.0 atm. calculate the work associated wi

th this process
Physics
2 answers:
Elenna [48]2 years ago
5 0

Answer:

The work is 9.99J.

Explanation:

The work associated with a process is given by the following formula:

W = P*\Delta

In which W is the work, P is the pressure and \Delta is the difference of volume.

In this problem, we have that:

The volume of an ideal gas is decreased from 5.0 l to 5.0 ml. Each ml has 0.001l. So the volume of the ideal gas is decreased from 5.0l to 0.005l. So

\Delta = 4.995.

Constant pressure of 2.0 atm. So P = 2.

W = P*\Delta

W = 2*4.995

W = 9.99

The work is 9.99J.

Fynjy0 [20]2 years ago
3 0
5 ml = 0,05l
Work = Pressure * (Volume difference)
W= 2 * (V2 - V1)
W= 2 * (4,95)
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An ideal gas is contained in a vessel at 300 K. The temperature of the gas is then increased to 900 K. (i) By what factor does t
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The question is missing some parts. Here is the complete question.

An ideal gas is contained in a vessel at 300K. The temperature of the gas is then increased to 900K.

(i) By what factor does the average kinetic energy of the molecules change, (a) a factor of 9, (b) a factor of 3, (c) a factor of \sqrt{3}, (d) a factor of 1, or (e) a factor of \frac{1}{3}?

Using the same choices in part (i), by what factor does each of the following change: (ii) the rms molecular speed of the molecules, (iii) the average momentum change that one molecule undergoes in a colision with one particular wall, (iv) the rate of collisions of molecules with walls, and (v) the pressure of the gas.

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              (ii) (c) a factor of \sqrt{3};

              (iii) (c) a factor of \sqrt{3};

             (iv) (c) a factor of \sqrt{3};

              (v) (e) a factor of 3;

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KE=\frac{3}{2}nRT

where

n is mols

R is constant of gas

T is temperature in Kelvin

As you can see, kinetic energy and temperature are directly proportional: when tem perature increases, so does energy.

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For temperature and energy, the factor of change is 3.

(ii) Rms is root mean square velocity and is defined as

V_{rms}=\sqrt{\frac{3k_{B}T}{m} }

Calculating velocity for each temperature:

For 300K:

V_{rms1}=\sqrt{\frac{3k_{B}300}{m} }

V_{rms1}=30\sqrt{\frac{k_{B}}{m} }

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V_{rms2}=\sqrt{\frac{3k_{B}900}{m} }

V_{rms2}=30\sqrt{3}\sqrt{\frac{k_{B}}{m} }

Comparing both veolcities:

\frac{V_{rms2}}{V_{rms1}}= (30\sqrt{3}\sqrt{\frac{k_{B}}{m} }) .\frac{1}{30} \sqrt{\frac{m}{k_{B}} }

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For rms, factor of change is \sqrt{3}

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q = m.v

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