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NARA [144]
4 years ago
5

Which substance(s) have no fixed shape and no fixed volume?

Engineering
2 answers:
Black_prince [1.1K]4 years ago
5 0

Answer:

I think gas substances

stiks02 [169]4 years ago
4 0

Answer:

Gas substances have no fixed shape or volume.

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Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is
Leona [35]

Answer:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

dQ = dU - dW

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) \int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW

As per work definition:

dW = F*dr

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

dW = PA*dx

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

dW = - P*dV

So the third integral in equation (1) is:

\int\limits^1_2 {- P} \, dV

Considering the gas as ideal, the pressure can be calculated as P = \frac{n*R*T}{V}, so:

\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Replacion this and solving equation (1) between state 1 and 2:

\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}

The internal energy depends only on the temperature of the gas, so there is no internal energy change U_{2} - U_{1} = 0, so the heat exchanged to the system equals the work done by the system:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

4 0
4 years ago
By adding "-once", one can form the noun form of the word "organize" is that true or false?​
denpristay [2]

Answer:

<h2>False </h2>

Explanation:

The noun form of organize is just adding letter r

7 0
3 years ago
Read 2 more answers
This is it dont anwser this is for my other account
Nezavi [6.7K]

Answer:

thanks for the poiunts

Explanation:

6 0
3 years ago
There are two types of cellular phones, handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles. P
nexus9112 [7]

Answer:

A) P(W) = 0.5

B) P(MF) = 0.3

C) P(H) = 0.6

Explanation:

We are told that there are two types of cellular phones which are handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles.

Also, Phone calls can be classified by the traveling speed of the user as fast (F) or slow (W).

Thus, the sample space is combination of types and classification we are given and it is written as;

S = {HF, HW, MF, MW}

A) Now, phones can either be fast(F) or slow(W). Thus, we can write;

P(F) + P(W) = 1

We are given P(F) = 0.5

Thus;

0.5 + P(W) = 1

P(W) = 1 - 0.5

P(W) = 0.5

B) Now, from the problem statement, a phone call can either be made with a handheld(H) or mobile(M). Thus the sample space partition is {H, M} and we can express as;

P(H ∩ F) + P(M ∩ F) = P(F)

We are given P[F] = 0.5 and P[HF] = 0.2.

P(H ∩ F) is same as P[HF]

Also, P(M ∩ F) is same as P(MF)

Thus;

0.2 + P(MF) = 0.5

P(MF) = 0.5 - 0.2

P(MF) = 0.3

C) Similarly, mobile Phone calls can either be fast or slow. It means the sample space partition is {F, W}

Thus;

P(M) = P(MW) + P(MF)

P(M) = 0.1 + 0.3

P(M) = 0.4

Now, since cellular phones can either be handheld(H) or Mobile(M), then we can say;

P(H) + P(M) = 1

P(H) + 0.4 = 1

P(H) = 1 - 0.4

P(H) = 0.6

5 0
3 years ago
How deep is a 6ft hole?​
Pavlova-9 [17]

Answer:

I know this sounds quite deep but it is as deep as a grave

Explanation:

It's reality

3 0
2 years ago
Read 2 more answers
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