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Flauer [41]
3 years ago
8

When a certain gas under a pressure of 5.00 × 106 Pa at 25.0°C is allowed to expand to 3.00 times its original volume, its final

pressure is 1.07 × 106 Pa. What is its final temperature?
Physics
1 answer:
12345 [234]3 years ago
4 0

Answer:

191.316 K or -81.684 °C

Explanation:

From general gas law,

P₁V₁/T₁ = P₂V₂/T₂ ................ Equation 1

Where P₁ = Initial pressure, V₁ = Initial volume, T₁ = Initial temperature, P₂ = Final pressure, V₂ = Final volume, T₂ = Final Temperature.

Make T₂ the subject of the equation.

T₂ = P₂V₂T₁/P₁ V₁ ............... Equation 2

Given: P₁ = 5.00×10⁶ Pa, T₁ = 25.0°C = 298 K, P₂ = 1.07×10⁶.

Let: V₁ = y cm³, V₂ = 3y cm³

Substitute into equation 2,

T₂ = (1.07×10⁶×298×3y)/(5.00×10⁶×y)

T₂ =191.316 K.

Hence the final temperature = 191.316 K or -81.684 °C

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Explanation:

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Answer:

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First we need to define an Frame of reference. Lets put the x axis unit vector \hat{i} pointing east,  with the y axis unit vector \hat{j} pointing south, so the positive angle is south of east. For this, we got for the first force:

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as is pointing north, and for the second force:

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as is pointing west.

Now, our third force will be:

\vec{F}_3  = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})

\vec{F}_3  =  83.7 \ N \ \hat{j}  + 59.9 \ N \ \hat{i}

\vec{F}_3  =  (59.9 \ N , 83.7 \ N )

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

|\vec{R}| = \sqrt{R_x^2 + R_y^2}

|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}

|\vec{F}_3| = 102.92 \ N

this is the magnitude.

To find the direction, we can use:

\theta = arctan(\frac{F_{3_y}}{F_{3_x}})

\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })

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