All categories

2 years ago

Describe the relationship between volume and temperature of an ideal gas

Chemistry

1 answer:

kvv77 [185]2 years ago

8 0

Answer:

Explanation:

Here, we want to describe the relationship between the volume and temperature of an ideal gas

This relationship is defined by Charles' law

From this law, we know that the volume of a given mass of gas is directly proportional to its temperature at a fixed pressure

What this means is that as long as the pressure remains unchanged, when the volume increases, the temperature increases, and when the volume decreases, the temperature decreases

These can be represented by the mathematical formula below:

$\frac{V_1}{T_1}\text{ = }\frac{V_2}{T_2}$

You might be interested in

Hydrogen peroxide decomposes spontaneously to yield water and oxygen gas according to the following reaction equation. 2H202(aq)

inn [45]

Answer : The temperature for non-catalyzed reaction needed will be 456 K

Explanation :

Activation energy : The energy required to initiate the reaction is known as activation energy.

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

Since, the rate for both the reaction are equal.

$K_1=K_2$

$A\times e^{\frac{-Ea_1}{RT_1}}=A\times e^{\frac{-Ea_2}{RT_2}}$

$\frac{Ea_1}{T_1}=\frac{Ea_2}{T_2}$ ..........(1)

where,

$Ea_1$ = activation energy for non-catalyzed reaction = 75 kJ/mol

$Ea_2$ = activation energy for catalyzed reaction = 49 kJ/mol

$T_1$ = temperature for non-catalyzed reaction = ?

$T_2$ = temperature for catalyzed reaction = $25^oC=273+25=298K$

Now put all the given values in the above formula 1, we get:

$\frac{Ea_1}{T_1}=\frac{Ea_2}{T_2}$

$\frac{75kJ/mol}{T_1}=\frac{49kJ/mol}{298K}$

$T_1=456K$

Therefore, the temperature for non-catalyzed reaction needed will be 456 K

3 0

3 years ago

How many L of 3.0 M H2SO4 solution can be prepared by using 100.0 mL OF 18 M H2SO4?

max2010maxim [7]

Answer: A volume of 600 mL of 3.0 M $H_{2}SO_{4}$ solution can be prepared by using 100.0 mL OF 18 M $H_{2}SO_{4}$ .

Explanation:

Given: $V_{1}$ = ?, $M_{1} = 3.0 M\\$

$V_{2} = 100.0 mL$ , $M_{2} = 18 M$

Formula used to calculate the volume is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\3.0 M \times V_{1} = 18 M \times 100.0 mL\\V_{1} = \frac{18 M \times 100.0 mL}{3.0M}\\= 600 mL$

Thus, we can conclude that a volume of 600 mL of 3.0 M $H_{2}SO_{4}$ solution can be prepared by using 100.0 mL OF 18 M $H_{2}SO_{4}$ .

6 0

3 years ago

Why can the mass of beta particles be ignored in nuclear reactions?

Harlamova29_29 [7]

Because their mass is so low that it doesn't really matter to us.

8 0

3 years ago

Jason is preparing curry from an Indian cookbook his British brother-in-law gee him for his birthday. Unfortunately, the British

Marina CMI [18]

Answer:

0.99 pounds

2,000 oz

0.423 cups

Explanation:

In order to convert this units we need to look up their equivalences.

1 g equals 0,0022 pounds approximately

Then we need to cross-multiply:

$\frac{1 g}{0.0022 pounds} = \frac{450 g}{ Xpounds} \\x = \frac{450 g x 0.0022 pounds}{1 g}$

450 g equals 0.99 pounds approximately

We can do the same calculation for the other 2 ingredients

1 g equals 16 oz

Then (125 g x 16 oz) / 1 oz = 2,000 oz (or 500 4 oz)

1 cup equals 236.59 ml

Then (100 ml x 1 cup)/ 236.59 = 0.423 cups

6 0

3 years ago

How fast should you heat the melting point sample?

hichkok12 [17]

Answer:

Option a.

Explanation:

The transformation of a solid into liquid is a slow process, hence a rate of heating too fast, near the melting point of the sample, will not give the right time to the crystals of the sample to absorb the heat and to melt in the outside and the inside, leading to wrong results in all cases. If we heat to fast, the melting point range will be too broad and will be misleading, resulting in values of the range more hight than the theoric ones.

A rate of 1 °C/min or 2 °C/min is the most appropriate to approach the melting point of the sample. Since it is too slow, from preventing the experiment taking forever it is recommended to start the experiment at a high heating rate until it reaches 20 °C below the melting point, and then, turn the heating rate down to 1 °C/min or 2 °C/min.

Therefore, the correct answer is a: near the melting point of the sample, we must heat slowly until it reach the expected temperature.

I hope it helps you!

7 0

3 years ago