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2 years ago

5) What is the pressure of a 3 moles of oxygen in a 5 liter container at 300

Mathematics

1 answer:

blondinia [14]2 years ago

7 0

Answer:

1496.52 Pa, 14.76 atm

Step-by-step explanation:

This question can be solved using ideal gas law

according to which

pv = nrt

where p is the pressure of gas

v is volume of gas

n is amount gas in moles

R is ideal gas constant whose value is 8.314 J/mol

T is absolute temperature of gas on Kelvin scale

______________________________

given

v = 5 liter

n = 3

T = 300 kelvin

R = 8.314 J/mol

Let the pressure be p

Substituting the above value in ideal gas law equation we have

p*5 = 3*8.314*300

p = 7482.6/5 = 1496.52 Pa

This pressure is in pascal while option is in atm

we know 1 Pascal is equal to 0.0000098692 atm

we know that 1 Atm = 101325 Pa

thus, 1496.52 is equal to 0.0000098692*1496.52 atm

1496.52 is equal to 14.76 atm

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see the explanation

Step-by-step explanation:

see the attached figure to better understand the problem

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2 years ago

What is the best approximation of the projection of (5,-1) onto (2,6)?

Hatshy [7]

Answer:

Hence, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\sqrt{10} }{5}$ , and the vector projection of $\vec a$ onto $\vec b$ = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$ .

Step-by-step explanation:

We have given two points $(5, -1)$ and $(2, 6)$ .

Let, $\vec a=5\hat {i}-\hat {j}$ and $\vec b= 2\hat {i}+6\hat{j}$ .

and we have calculate the projection of $\vec a$ onto $\vec b$ .

Now,

For the calculation of projection, first we need to calculate the dot product of $\vec a$ and $\vec b$ .

$\vec a.\vec b=(5\hat {i}-\hat{j}).(2\hat{i}+6\hat{j})$

$=10-6$

$=4$

then, we have to calculate the magnitude of $\vec b$ .

$\mid {\vec {b}}\mid$ = $\sqrt{2^{2}+6^{2} }$ = $\sqrt{40}$ = $2\sqrt{10}$ .

Now, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\vec a.\vec b}{\mid b\mid}$

= $\frac{4}{2\sqrt{10} }$ $\frac{2}{\sqrt{10} } \times\frac{\sqrt{10} }{\sqrt{10} } =\frac{2\sqrt{10} }{10} = \frac{\sqrt{10} }{5}$

and the vector projection of $\vec a$ onto $\vec b$ = $\frac{\vec a. \vec b}{\mid\vec b \mid^{2} } . \vec b$

= $\frac{4}{40} . (2\hat i+ 6\hat j)$

= $\frac{1}{5} \hat i+\frac{3}{5} \hat j$

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