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salantis [7]
3 years ago
11

A 0.030-m3 container is initially evacuated. Then, 4.0 g of water is placed in the container, and, after some time, all the wate

r evaporates. If the temperature of the water vapor is 388 K, what is its pressure?
Physics
1 answer:
stellarik [79]3 years ago
4 0

Answer:

23895.05 Pa

Explanation:

V = 0.03 m³

mass, m = 4 g

Temperature, T = 388 K

Let P be the pressure.

number of moles, n = 4 / 18 = 0.22

Use ideal gas equation

PV = n RT

P x 0.03 = 0.22 x 8.314 x 388

P = 23895.05 Pa

Thus, the pressure is 23895.05 Pa.

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Block A weighs 14 lb and block B weighs 5 lb If B is moving downward with a velocity (vB)1 = 3 ft/s at t = 0, determine the velo
kolezko [41]

Velocity, va2 = 10.5 ft/s

<u>Explanation:</u>

From the figure:

Length of the cable = Sa + 2Sb = l

∴ vₐ = -2vb

Applying the principle of Impulse and momentum in x-direction

mv_x_1 + \int\limits^t_t {F_x} \, dt = mv_x_2

Limit is t1 to t2

-(\frac{14}{32.2}) (2) (3) - T = \frac{14}{32.2} v_y_2                           -(1)

Applying the principle of Impulse and momentum in y-direction

mv_y_1 + \int\limits^t_t {F_y} \, dt = mv_y_2

Limit is t1 to t2

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T = 1.6lb

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