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Vilka [71]
4 years ago
10

4.39 moles of gas in a box has a pressure of 2.25 atm at temperature of 385K. What is the volume of the box?

Physics
2 answers:
stira [4]4 years ago
6 0

Answer:

Volume of the container = 0.0618 m³

Explanation:

Assuming that the gas is ideal, we can apply ideal gas equation:

PV=nRT

where P = pressure of the gas

          V = volume occupied by the gas

          n =  number of moles of gas

          R = Universal gas constant = 8.314 J/mol-K

          T = Temperature of the gas

Here we have to find volume.

Given: P=2.25 atm T=385 K n=4.39 moles

Putting the above values in the ideal gas equation, we get:

V=\frac{nRT}{P}=\frac{4.39\times8.314\times385}{2.25\times1.01\times10^{5}}=0.0618\ m^3

Volume of the gas = volume of the container in which it is filled = 0.0618 m³

bazaltina [42]4 years ago
5 0

Answer:

0.0619 m^3

Explanation

number of moles = n = 4.39 mol

pressure = P = 2.25 atm =2.25×1.01×10^5 Pa= 2.27×10^5 Pa

Molar gas constant =R = 8.31 J/(mol K)

Temperature T= 385K

volume of gas = V =?

BY GENERAL GAS LAW WE HAVE

PV = nRT

or V = nRT/P

or V = (4.39×8.31×385)/(2.27×10^5)

V = 0.0618728

V =  0.0619 m^3

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A car with mass 950 kg and a speed of 16 m/s approaches an intersection. A 1300 kg minivan traveling at 21 m/s is heading for th
Alex73 [517]

Answer:

V_f = 13.8863 \angle 60.89\°

Explanation:

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m_1 = 950Kg\\v_1 = 16m/s \\m_2 =1300Kg\\v_2 = 21m/s

We have all the values to apply the law of linear momentum, however, it is necessary to define the two lines in which the study will be carried out. Being an intersection the vehicle of mass m_1 approaches through the X axis, while the vehicle of mass m_2 approaches by the y axis. In the collision equation on the X axis, we despise the velocity of object 2, since it does not come in this direction.

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For the particular case on the Y axis, we do the same with the speed of object 1.

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By taking a final velocity as a component, we can obtain the angle between the two by relating the equations through the tangent

Tan\theta = \frac{m_2v_2}{m_1v_1}\\Tan\theta = \frac{1300*21}{950*16}\\\theta = tan^{-1}(1.7960)\\\theta = 60.89\°

Replacing in any of the two functions, given above, we will find the final speed after the collision,

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V_f= \frac{(950)(16)}{(950+1300)cos(60.89)}

V_f = 13.8863 \angle 60.89\°

8 0
3 years ago
Capacitances of 10uF and 20uF are connected in parallel,
jarptica [38.1K]

Answer:

The equivalent capacitance will be 15\mu F  

Explanation:

We have given two capacitance C_=10\mu F\ and\ C_2=20\mu F

They are connected in parallel

So equivalent capacitance C=C_1+C_2=10+20=30\mu F

This equivalent capacitance is now connected in series with 30\mu F

In series combination of capacitors the equivalent capacitance is given by \frac{1}{C}=\frac{1}{30}+\frac{1}{30}

C=\frac{30}{2}=15\mu F

So the equivalent capacitance will be 15\mu F  

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