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valentina_108 [34]

3 years ago

13

You want to store 138 g of gas in a 13.8-l tank at room temperature (25 °c). calculate the pressure the gas would have using the

ideal gas law and the van der waals equation

1 answer:

NikAS [45]3 years ago

6 0

MMMMMMMMMMMMMMMMMMMMMMMMMMMMMM

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At the end of the adiabatic expansion, the gas fills a new volume V₁, where V₁ > V₀. Find W, the work done by the gas on the

tino4ka555 [31]

Answer:

$W=\frac{p_0V_0-p_1V_1}{\gamma-1}$

Explanation:

An adiabatic process refers to one where there is no exchange of heat.

The equation of state of an adiabatic process is given by,

$pV^{\gamma}=k$

where,

$p$ = pressure

$V$ = volume

$\gamma=\frac{C_p}{C_V}$

$k$ = constant

Therefore, work done by the gas during expansion is,

$W=\int\limits^{V_1}_{V_0} {p} \, dV$

$=k\int\limits^{V_1}_{V_0} {V^{-\gamma}} \, dV$

$=\frac{k}{\gamma -1} (V_0^{1-\gamma}-V_1^{1-\gamma})\\$

(using $pV^{\gamma}=k$ )

$=\frac{p_0V_0-p_1V_1}{\gamma-1}$

4 0

3 years ago

A radar station detects an airplane approaching directly from the east. At first observation, the range to the plane is 375.0 m

Gekata [30.6K]

Answer:

819.78 m

Explanation:

<u>Given:</u>

OA = range of initial position of the airplane from the point of observation = 375 m
OB = range of the final position of the airplane from the point of observation = 797 m
$\theta$ = angle of the initial position vector from the observation point = $43^\circ$
$\alpha$ = angle of the final position vector from the observation point = $123^\circ$
$\vec{AB}$ = displacement vector from initial position to the final position

A diagram has been attached with the solution in order to clearly show the position of the plane.

$\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m$

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

$\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} = (-708.34\ \hat{i}+412.67\ \hat{j})\ m$

The magnitude of the displacement vector = $\sqrt{(-708.34)^2+(412.67)^2}\ m = 819.78\ m$

Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.

8 0

3 years ago

Question 2 (1 point)

sveticcg [70]

according to the facts it should be a numerous amount of wrong answers so ignore dis

7 0

3 years ago

What are discretionary calories?

AleksAgata [21]

<span>is the excess </span>calories<span> you can have once you get your required nutrient </span><span>
those excess calories can come from higher forms of fat such as milk cheese and meat and sugary items such as syrup butter and any sauce those calories can also come from pop candy and alcohol </span>

8 0

3 years ago

At a particular instant the magnitude of the gravitational force exerted by a planet on one of its moons is 7 × 1021 N. Collapse

9966 [12]

Answer:

Fg = 4.2*10²² N

Explanation:

The gravitational force between any two masses, provided that can be approximated by point masses (comparing their diameters with the distance between them), obeys the Newton's Universal Law of Gravitation, which states that the force (always attractive) is proportional to the product of the masses and inversely proportional to the square of the distance between them (this as a consequence of our Universe being three-dimensional), as follows:

$Fg =\frac{G*m1*m2}{r^{2}}$

So, if one of the masses increases 6 times, the force between them will be directly 6 times larger, so the new magnitude of the force will be as follows:

Fg₂ = Fg₁*6 = 7*10²¹ N* 6 = 4.2*10²² N

7 0

3 years ago