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stiks02 [169]
3 years ago
15

Describe a scenario using the combined gas law in which changes occur in the pressure and temperature of an enclosed gas but the

volume does not change.
Chemistry
2 answers:
Damm [24]3 years ago
7 0

If the pressure and the temperature of an enclosed gas are doubled, the volume does not change.

The combined gas law describes the change in the volume of an ideal gas when the pressure and the temperature are changed simultaneously. The mathematical expression is:

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where,

  • P: pressure
  • V: volume
  • T: temperature
  • 1: initial state
  • 2: final state

Let's imagine an ideal gas that occupies a certain volume (V₁) at a certain pressure (P₁) and temperature (T₁). If the pressure were doubled (P₂ = 2 P₁) and the temperature were doubled (T₂ = 2 T₁), the new volume (V₂) would be:

V_2 = \frac{P_1V_1T_2}{T_1P_2}  = \frac{P_1V_1(2T_1)}{T_1(2P_1)}  = V_1

If the pressure and the temperature of an enclosed gas are doubled, the volume does not change.

Learn more: brainly.com/question/13154969

zimovet [89]3 years ago
4 0

Answer:

There are 4 general laws that relate the 4 basic characteristic properties of gases to each other. Each law is titled by its discoverer. While it is important to understand the relationships covered by each law, knowing the originator is not as important and will be rendered redundant once the combined gas law is introduced. So concentrate on understanding the relationships rather than memorizing the names.

Charles' Law- gives the relationship between volume and temperature if the pressure and the amount of gas are held constant:

1) If the Kelvin temperature of a gas is increased, the volume of the gas increases. (P, n Constant)

2) If the Kelvin temperature of a gas is decreased, the volume of the gas decreases. (P, n Constant)

This means that the volume of a gas is directly proportional to its Kelvin temperature. Think of it this way, if you increase the volume of a gas and must keep the pressure constant the only way to achieve this is for the temperature of the gas to increase as well.

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Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: spe
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Answer:

The amount of heat to absorb is 6,261 J

Explanation:

Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.

The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.

So:

  • Heat required to raise the temperature of ice from -20 °C to 0 °C

Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).

In this case, m= 10 g, specific heat of the ice= 2.1 \frac{J}{g*C} and ΔT=0 C - (-20 C)= 20 C

Replacing: Q= 10 g*2.1 \frac{J}{g*C} *20 C and solving: Q=420 J

  • Heat required to convert 0 °C ice to 0 °C water

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

In this case, being 1 mol of water= 18 grams: Q= 10 g*6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}= 3.333 kJ= 3,333 J (being kJ=1,000 J)

  • Heat required to raise the temperature of water from 0 °C to 60 °C

In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 \frac{J}{g*C} and ΔT=60 C - (0 C)= 60 C

Replacing: Q= 10 g*4.18 \frac{J}{g*C} *60 C and solving: Q=2,508 J

Finally, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

<u><em> The amount of heat to absorb is 6,261 J</em></u>

<u><em></em></u>

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