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3 years ago

What must happen for water to boil at a given pressure?

Chemistry

2 answers:

Mademuasel [1]3 years ago

7 0

Answer:

B. The water must gain thermal energy from its surroundings.

Explanation:

Boiling is the process whereby a substance is changed from a liquid state to a gaseous state due to the application of heat. For boiling to be achieved, a liquid substance must reach or exceed its boiling point, which is the specific temperature at which a specific liquid boils.

For this change of state i.e. from liquid to gas, to occur, atoms or molecules of the liquid must gain kinetic energy and move faster. However, this kinetic energy cannot be attained without a rise in temperature, which causes the molecules to gain THERMAL ENERGY or heat energy and move fast. Thermal energy is the energy that arises from a hot substance.

Therefore, for water to boil at a given pressure, it must gain thermal energy from its external surroundings, which causes its molecules to have an increased kinetic energy and move faster. This fast movement of the molecules causes the transition of phases (from liquid to gas).

BaLLatris [955]3 years ago

7 0

The water must give off thermal energy to its surroundings.

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almond37 [142]

Answer:

C. 12.8 liters.

Explanation:

The Standard Temperature and Pressure (STP) of a gas are 273.15 K and 100 kilopascals. From Avogadro's Law, a mole of carbon dioxide contains $6.022 \times 10^{23}$ molecules. If we suppose that carbon dioxide behaves ideally, then the equation of state for ideal gas is:

$P\cdot V = n\cdot R_{u}\cdot T$ (1)

$P\cdot V = \frac{r\cdot R_{u}\cdot T}{N_{A}}$ (1b)

Where:

$P$ - Pressure, measured in pascals.

$V$ - Volume, measured in liters.

$r$ - Amount of molecules, no unit.

$N_{A}$ - Avogadro's number, no unit.

$R_{u}$ - Ideal gas constant, measured in pascal-liters per mole-Kelvin.

$T$ - Temperature, measured in Kelvin.

If we know that $P = 100000\,Pa$ , $r = 3.44\times 10^{23}$ , $N_{A} = 6.022\times 10^{23}$ , $T = 273.15\,K$ and $R_{u} = 8.314\times 10^{3}\,\frac{L\cdot Pa}{mol\cdot K}$ , then the volume of carbon dioxide at STP is:

$V = \frac{r\cdot R_{u}\cdot T}{N_{A}\cdot P}$

$V = \frac{(3.44\times 10^{23})\cdot \left(8.314\times 10^{3}\,\frac{L\cdot Pa}{mol\cdot K} \right)\cdot (273.15\,K)}{(6.022\times 10^{23})\cdot (100000\,Pa)}$

$V = 12.972\,L$

Therefore, the correct answer is C.

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Explanation:

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