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VLD [36.1K]
3 years ago
8

A gas sample is heated from -20.0°C to 57.0°C and the volume is increased from 2.00 L to 4.50 L. If the initial pressure is 0.14

0 atm, what is the final pressure? Select one: a. 0.0477 atm b. –0.177 atm c. 0.411 atm d. 0.242 atm e. 0.0811 atm
Physics
1 answer:
Svetach [21]3 years ago
5 0

Answer:

The answer is e.

Explanation:

We take:

T_{1}=253K

V_{1}=2.00 l

P_{1}=0.14atm

V_{2}=4.50 l

T_{2}=330K

Taking the gas as an ideal gas, we can use the ideal gas law:

\frac{PV}{nRT}

⇒ n=\frac{P_{1}V_{1}}{RT_{1}} ⇒ n=0.013mol

Then:

P_{2}=\frac{nRT_{2}}{V_{2}} ⇒ P_{2}=0.0811 atm

Taking R=0.08205atmL/molK.

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E) Thermal energy is released during
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Answer:

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Explanation:

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Q_{net,in} - W_{net, out} = \Delta U (1)

Where:

Q_{net, in} - Net input heat, measured in joules.

W_{net, out} - Net output work, measured in joules.

\Delta U - Change in thermal energy, measured in joules.

Please notice that work comprises all kind of work (i.e. mechanical, electric, magnetic), whereas heat comprises all heat interactions including chemical and radioactive phenomena.

If thermal energy is released, then \Delta U < 0, which is caused by three scenarios:

(i) Q_{net,in} < 0, W_{net, out} < 0, |Q_{net,in}|>|W_{net,out}|

(ii) Q_{net, in} > 0, W_{net,out} > 0, |Q_{net,in}|

(iii) Q_{net,in}< 0, W_{net, out}>0

In the case Q_{net,in} > 0, W_{net, out}, the thermal energy of the system is increased. Therefore, thermal energy is released during some energy conversions. Answer: True

f) A liquid solidifies when temperature goes below point of fusion, meaning a realease of heat with no work interactions. That is:

Q_{net, in} = \Delta U, Q_{net, in} < 0 (2)

If Q_{net, in} < 0, then  \Delta U < 0. Then, if a liquid absorbs heat energy, then thermal energy is increase and the liquid does not solidifies. Answer: False.

7 0
2 years ago
So what would the answer for D be?
Mademuasel [1]
For what, exactly? XD
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