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son4ous [18]
3 years ago
6

A sample of nitrogen gas is inside a sealed container. The container is slowly compressed, while the temperature is kept constan

t. This is a(n) ________ process. A sample of nitrogen gas is inside a sealed container. The container is slowly compressed, while the temperature is kept constant. This is a(n) ________ process.
a. adiabatic
b. constant-volume
c. isobaric
d. isothermal
Chemistry
1 answer:
Mkey [24]3 years ago
8 0

Answer:

The correct option is  d isothermal

Explanation:

Step 1: Data given

An adiabatic process is a process in which no <u>heat </u>is gained or lost by the system.

A constant- volume process, also called an isochoric process is a process in which the <u>volume is held constant</u>, meaning that the work done by the system will be zero.

An Isobaric process is a thermodynamic process taking place at <u>constant pressure</u>

<u />

An isothermal process is a thermodynamic process, in which the<u> temperature of the system remains constant </u>

Step 2: This situation

The container is slowly compressed, while the temperature is kept constant.

This is  an isothermal process.

The correct option is  d isothermal

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Explanation:

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On a recent trip to the Fort Worth Zoo, Trevor's parents bought him a helium-filled balloon. At the beginning of the day, the sk
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Answer:

1. The new volume is 14L

2. The volume decreased

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 14.2 L

Initial pressure (P1) = 102.5 kPa

Initial temperature (T1) = 33°C

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Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below:

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T1 = 33°C = 33°C + 273 = 306K

T2 = 24°C = 24°C + 273 = 297K

Step 3:

Determination of the new volume.

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P1V1/T1 = P2V2/T2

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Cross multiply to express in linear form as shown below:

306 x 100.9 x V2 = 102.5 x14.2 x297

Divide both side by 306 x 100.9

V2 = (102.5x14.2x297)/(306 x 100.9)

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Determination of the change in volume. This is illustrated below

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Change in volume (ΔV) = Final volume (V2) - Initial volume (V1)

ΔV = V2 - V1

ΔV = 14 - 14.2

ΔV = - 0.2L

Since the change in volume is negative, it means there is a decrease in the volume.

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