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3 years ago

When a sample of a gas is heated in a sealed, rigid container from 200. K to 400. K, the pressure exerted by the gas is:

1 answer:

5 0

When a sample of a gas is heated in a sealed, rigid container from 200 degree Kelvin to 400 degree Kelvin, the pressure exerted by the gas is increased by a factor 2. Heating any gas actually increases the volume of the gas within a container. As the temperature of the gas rises, the molecules of the gas start moving faster and start striking the walls of the container in which it is kept with more force. The volume of the container tries to expand to accommodate the fast colliding molecules of the gas.

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A student hypothesized that the green color of plant leaves is related to exposure to sunlight and tested this hypothesis by pla

kkurt [141]

The hypothesis is that
the green coloration of plants is related to sunlight exposure of leaves
The hypothesis is tested by keeping a dark green plant inside a dark closet
The result is that the green coloration faded
The conclusion must be that exposure to sunlight affects the green color of the plants

8 0

3 years ago

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A vessel of volume 22.4 dm3 contains 20 mol h2 and 1 mol n2 ad 273.15 k initially. All of the nitrogen reacted with sufficient h

NikAS [45]

Nitrogen combine with hydrogen to produce ammonia $\text{NH}_3$ at a $1:3:2$ ratio:

$\text{N}_2 \; (g) + 3 \; \text{H}_2 \; (g) \leftrightharpoons 2\; \text{NH}_3 \; (g)$

Assuming that the reaction has indeed proceeded to completion- with all nitrogen used up as the question has indicated. $3 \; \text{mol}$ of hydrogen gas would have been consumed while $2 \; \text{mol}$ of ammonia would have been produced. The final mixture would therefore contain

$17 \; \text{mol}$ of $\text{H}_2 \; (g)$ and
$2 \; \text{mol}$ of $\text{NH}_3 \; (g)$

Apply the ideal gas law to find the total pressure inside the container and the respective partial pressure of hydrogen and ammonia:

$\begin{array}{lll} P(\text{container}) &= & n \cdot R \cdot T / V \\ & = & (17 + 2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 1.926 \times 10^{3} \; \text{kPa} \end{array}$
$\begin{array}{lll} P(\text{H}_2) &= & n \cdot R \cdot T / V \\ & = & (17) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 1.723 \times 10^{3} \; \text{kPa} \end{array}$
$\begin{array}{lll} P(\text{NH}_3) &= & n \cdot R \cdot T / V \\ & = & (2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 2.037 \times 10^{2} \; \text{kPa} \end{array}$

6 0

3 years ago

Calcutale Grxn for the following equation at 25°C: 4KClO3(s) → 3KClO4(s) KCl(s)

Step2247 [10]

Answer:

-133.2 kJ

Explanation:

Let's consider the following balanced equation.

4 KClO₃(s) → 3 KClO₄(s) + KCl(s)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression.

ΔG°rxn = 3 mol × ΔG°f(KClO₄(s)) + 1 mol × ΔG°f(KCl(s)) - 4 mol × ΔG°f(KClO₃(s))

ΔG°rxn = 3 mol × (-303.1 kJ/mol) + 1 mol × (-409.1 kJ/mol) - 4 mol × (-296.3 kJ/mol)

ΔG°rxn = -133.2 kJ

5 0

3 years ago

What are the properties of gas

Oksana_A [137]

Answer:

1) easy compressed

2) fills its container

3) far more space

Explanation:

5 0

3 years ago

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2. 2. Consider the following reaction occurring in a closed chemical system. Assume that this reaction is at equilibrium and tha

Alika [10]

1. This is a combustion reaction.

Combustion reactions can happen with the presence of O₂ gas. O₂ reacts with another element or compound and oxidize it. Here ethanol reacts with O₂ and produces CO₂ and H₂O as products. Combustion is also called as burning. 

2. Reaction will shift to right. 

If more CH₃CH₂OH is added to the system, then the amount of CH₃CH₂OH will increase. Then the equilibrium in the system will be broken. To make the equilibrium again, the added CH₃CH₂OH should be removed. To do that system will consume more CH₃CH₂OH to make products which helps to decrease the amount of ethanol. Hence, the reaction will shift to right.

3. The reaction will shift to right.

If the water is extracted from the system, the amount of water will decrease. That means the amount of products decrease. Then the system will try to gain equilibrium by increasing the water. To increase water the forward reaction should be enhanced. Hence, the reaction will shift to right.

4. The reaction will shift to right.

This is an exothermic reaction since it produces heat. If the produced heat is removed, then the system will be cold. To maintain the temperature, system has to increase the amount of heat produced. Then, the forward reaction should be enhanced. Hence, the reaction will shift to right.

5. The Le Chatelier's principle.

Le Chatelier's principle says if a condition changes in a system which was in an equilibrium state, the system will try to gain equilibrium by correcting the changed condition back to normal. Most of industries which make chemicals use this principle

3 0

3 years ago

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