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Temka [501]
3 years ago
11

The stopcock connecting a 2.14 L bulb containing oxygen gas at a pressure of 8.19 atm, and a 9.84 L bulb containing krypton gas

at a pressure of 2.65 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is atm.
what is the final pressure of the system in atm?
Physics
1 answer:
marshall27 [118]3 years ago
4 0

Answer : The final pressure of the system in atm is, 3.64 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

P_1V_1+P_2V_2=P_fV_f

where,

P_1 = first pressure = 8.19 atm

P_2 = second pressure = 2.65 atm

V_1 = first volume = 2.14 L

V_2 = second volume = 9.84 L

P_f = final pressure = ?

V_f = final volume = 2.14 L  + 9.84 L = 11.98 L

Now put all the given values in the above equation, we get:

8.19atm\times 2.14L+2.65atm\times 9.84L=P_f\times 11.98L

P_f=3.64atm

Therefore, the final pressure of the system in atm is, 3.64 atm

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Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
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Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

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M=3.66(10)^{33}kg (11) This is the mass of the star S

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