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4 years ago

Sebanyak 0,2 mol gas ideal berada dalam wadah yang volumenya 10 liter dan tekanan 1 atm . berapakah suhu gas tersebut

Physics

1 answer:

tamaranim1 [39]4 years ago

8 0

The temperature of the gas is about 600 K

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<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

$\large {\boxed {PV = nRT} }$

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

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<u><em>Question (Translation):</em></u>

<em>A total of 0.2 moles of ideal gas are in containers with volume of 10 liters and a pressure of 1 atm. What is the temperature of the gas?</em>

<u>Given:</u>

number of moles = n = 0.2 moles

volume of gas = V = 10 liters

pressure of gas = P = 1 atm

gas contant = R = 0.0821 L.atm/mol.K

<u>Unknown:</u>

temperature of gas = T = ?

<u>Solution:</u>

$PV = nRT$

$1 \times 10 = 0.2 \times 0.0821 \times T$

$10 = 0.01642 \times T$

$T = 10 \div 0.01642$

$T \approx 600 \texttt{ K}$

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<h3>Learn more</h3>

Minimum Coefficient of Static Friction : brainly.com/question/5884009
The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant , Liquid , Pressure

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And we can find the positions for the two times required like this:

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And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

Explanation:

The particle position is given by:

$z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0$

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In order to find the velocity we need to take the first derivate for the position function like this:

$z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

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And we can find the positions for the two times required like this:

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And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

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