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user100 [1]
2 years ago
13

What is the effect of the following change on the volume of 1 mol of an ideal gas? The initial pressure is 722 torr, the final p

ressure is 0.950 atm, the initial temperature is 32°F, and the final temperature is 273 K. A. Volume is unchanged B. Volume increases by about 10% C. volume decreases by about 92% D. volume increases by 8% E. volume increases by about 90%.
Physics
1 answer:
Ray Of Light [21]2 years ago
4 0

Answer:

A. Volume is unchanged

Explanation:

P_{i} = initial pressure of the gas = 722 torr = 96258.7 pa

P_{f} = final pressure of the gas = 0.950 atm = 96258.75 pa

T_{i} = initial temperature = 32 °F = 272.15 K

T_{f} = final temperature = 273 K

V_{i} = initial volume

V_{f} = final volume

Using the Equation

\frac{P_{i} V_{i}}{T_{i}} = \frac{P_{f} V_{f}}{T_{f}}

Inserting the values

\frac{(96958.7) V_{i}}{272.15} = \frac{(96958.75) V_{f}}{273}

V_{f} = (1.00312) V_{i}

Hence the volume is unchanged.

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2 years ago
1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra
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Answer:

1/2

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I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

F_r = mgsin\theta

F_r = \mu N

F_r = \mu mg cos\theta

Matching the two equation we have,

\mu N = \mu mg cos\theta

\mu = tan\theta

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\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1

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Frictionless surface

\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2

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Answer:

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