Calculate the answer to the correct

Two moles of helium gas initially at 438 K and 0.44 atm are compressed isothermally to 1.61 atm. Find the final volume of the ga

docker41 [41]

Answer:

44.64335 L

Explanation:

R = Gas constant = 8.314 J/mol K = 0.08205 L atm/mol K

P = Pressure

V = Volume

T = Temperature = 438 K

1 denotes initial

2 denotes final

From ideal gas law we have

$PV=nRT\\\Rightarrow V=\dfrac{nRT}{P}\\\Rightarrow V=\dfrac{2\times 0.08205\times 438}{0.44}\\\Rightarrow V=163.35409\ L$

So,

$P_1V_1=P_2V_2\\\Rightarrow V_2=\dfrac{P_1V_1}{P_2}\\\Rightarrow V_2=\dfrac{0.44\times 163.35409}{1.61}\\\Rightarrow V_2=44.64335\ L$

The volume of Helium is 44.64335 L

5 0

3 years ago

A cruise ship sails due south at 2.50 m/s while a coast guard patrol boat heads 19.0° north of west at 4.80 m/s. What are the x-

Nesterboy [21]

Answer:

Explanation:

We shall represent the velocity of cruise ship and coast guard petrol boat in vector form .

velocity of cruise ship

Vcs = - 2.5 j

Vpb = - 4.8 cos 19 i + 4.8 sin 19 j = - 4.54 i + 1.56 j

velocity of the cruise ship relative to the patrol boat

= Vcs - Vpb

= - 2.5 j - ( - 4.54 i + 1.56 j )

= - 2.5 j + 4.54 i - 1.56 j

= 2.04 i - 1.56 j .

x-component of the velocity of the cruise ship relative to the patrol boat

= 2.04 m /s

y-component of the velocity of the cruise ship relative to the patrol boat

= - 1.56 m /s .

7 0

2 years ago

A man of weight Wman is standing on the second floor and is pulling on a rope to lift a box of weight Wbox from the floor below.

slavikrds [6]

Answer:

Explanation:

See the attached figure . See the forces acting on man pulling up the box .

Man is stationary so net force acting on man is zero .

T + R = Wman

R is the reaction force of the ground of second floor .

R = Wman - T

3 0

3 years ago

The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr

WARRIOR [948]

In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
$y_n= \frac{n \lambda D}{a}$ (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
$\lambda$ is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem,
$D=37.0 cm=0.37 m$
$\lambda=530 nm=5.3 \cdot 10^{-7} m$
while the distance between the first and the fifth minima is
$y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m$ (2)

If we use the formula to rewrite $y_5, y_1$ , eq.(2) becomes
$\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m$
Which we can solve to find a, the width of the slit:
$a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}= 1.57 \cdot 10^{-3} m=1.57 mm$

7 0

3 years ago

INK]: A car is moving at 200 km/hr east. What is its velocity?

Greeley [361]

Answer:

200 km\h

or 0.621 mp\h its the same speed just different mesuarements

5 0

2 years ago