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anastassius [24]
3 years ago
15

Why do the air molecules inside a bicycle tire speed up as the temperature gets warmer?

Chemistry
2 answers:
liraira [26]3 years ago
6 0
The correct answer of the given question above would be option A. The air molecules inside a bicycle tire speed up as the temperature gets warmer because the heat is transferred to the molecules and gives them more kinetic energy. <span>When </span>heat<span> is added to a substance, the </span>molecules<span> and atoms vibrate </span>faster<span>. </span>
Mice21 [21]3 years ago
5 0

Answer: Option (a) is the correct answer.

Explanation:

When temperature of surrounding atmosphere around the tire increases then it will lead to increase in the kinetic energy of molecules.

Due to increase in kinetic energy of molecules there will be more number of collisions between the air molecules that are present inside the bicycle tire.

Hence, it means heat has been transferred from the surrounding atmosphere to the molecules inside the tire.

Thus, we can conclude that the air molecules inside a bicycle tire speed up as the temperature gets warmer because the heat is transferred to the molecules and gives them more kinetic energy.

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A gold nugget has a density of 38.6g/cm3 and a mass of 270.2. what is its volume?​
aivan3 [116]

Density is given by the equation D=m/V, were D is density, m is mass in grams, and V is volume in cubic centimeters.


In this problem, we have density and we have mass so we can plug into the equation and solve for V.

38.6=270.2/V

<em>*Multiply both sides by V*</em>

38.6V=270.2

<em>*Divide both sides by 38.6*</em>

V=7


The volume of the gold nugget is 7cm3.


Hope this helps!!

5 0
3 years ago
Read 2 more answers
At standard temperature and pressure, what volume does 0.5 mol of CO2 occupyAt standard temperature and pressure, what volume do
Zigmanuir [339]
<span>C. 11.2 L There are several different ways to solve this problem. You can look up the density of CO2 at STP and work from there with the molar mass of CO2, but the easiest is to assume that CO2 is an ideal gas and use the ideal gas properties. The key property is that a mole of an idea gas occupies 22.413962 liters. And since you have 0.5 moles, the gas you have will occupy half the volume which is 22.413962 * 0.5 = 11.20698 liters. And of the available choices, option "C. 11.2 L" is the closest match. Note: The figure of 22.413962 l/mole is using the pre 1982 definition of STP which is a temperature of 273.15 K and a pressure of 1 atmosphere (1.01325 x 10^5 pascals). Since 1982, the definition of STP has changed to a temperature of 273.15 K and a pressure of exactly 10^5 pascals. Because of this lower pressure, one mole of an ideal gas will have the higher volume of 22.710947 liters instead of the older value of 22.413962 liters.</span>
4 0
3 years ago
Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which
Drupady [299]

Answer:

Explanation:

From the information given:

Mass of carbon tetrachloride = 5 kg

Pressure = 1 bar

The given density for carbon tetrachloride = 1590 kg/m³

The specific heat of carbon tetrachloride =  0.84 kJ/kg K

From the composition, the initial volume of carbon tetrachloride will be:= \dfrac{5 \ kg }{1590 \ kg/m^3}

= 0.0031 m³

Suppose \beta is independent of temperature while pressure is constant;

Then:

The change in volume can be expressed as:

\int ^{V_2}_{V_1} \dfrac{dV}{V} =\int ^{T_2}_{T_1} \beta dT

In ( \dfrac{V_2}{V_1})  = \beta (T_2-T_1)

V_2 = V_1 \times exp (\beta (T_2-T_1))

V_2 = 0.0031 \ m^3  \times exp  (1.2 \times 10^{-3} \times 20)

V_2 = 0.003175 \ m^3

However; the workdone = -PdV

W = -1.01 \times 10^5 \ Pa \times ( 0.003175 m^3 - 0.0031 \ m^3)

W = - 7.6 J

The heat energy Q = Δ h

Q = mC_p(T_2-T_1)

Q = 5 kg \times 0.84 \ kJ/kg^0 C \times 20

Q = 84 kJ

The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;

ΔU = ΔQ + W

ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)

ΔU = 83.992 kJ

3 0
3 years ago
Simple question plzz help
PSYCHO15rus [73]

Answer:

a  

Explanation:

3 0
2 years ago
Read 2 more answers
A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is −0.3 kJ/K
Karolina [17]

Answer:

W = -120 KJ

Explanation:

Since the piston–cylinder assembly undergoes an isothermal process, then the temperature is constant.

Thus; T1 = T2 = 400K

change in entropy; ΔS = −0.3 kJ/K

Formula for change in entropy is written as;

ΔS = Q/T

Where Q is amount of heat transferred.

Thus;

Q = ΔS × T

Q = -0.3 × 400

Q = -120 KJ

From the first law of thermodynamics, we can find the workdone from;

Q = ΔU + W

Where;

ΔU is Change in the internal energy

W = Work done

Now, since it's an ideal gas model, the change in internal energy is expressed as;

ΔU = m•C_v•ΔT

Where;

m is mass

C_v is heat capacity at constant volume

ΔT is change in temperature

Now, since it's an isothermal process where temperature is constant, then;

ΔT = T2 - T1 = 0

Thus;

ΔU = m•C_v•ΔT = 0

ΔU = 0

From earlier;

Q = ΔU + W

Thus;

-120 = 0+ W

W = -120 KJ

8 0
3 years ago
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