Physics, 17.09.2019 20:00 Amazingpandakid

An automobile tire is inflated with air originally at 10.0°c and normal atmospheric pressure. during the process, the air is compressed to 30.0% of its original volume and the temperature is increased to 46.0°c. (a) what is the pressure in the tire (absolute)? pa (b) after the car is driven at high speed, the tire's air temperature rises to 85.0°c and the tire's interior volume increases by 2.00%. what is the new pressure in the tire (absolute)? pa

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Answer from: pollywallythecat

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

Given:

= The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$ T_2

= The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$ T_3

= The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$ V_1

= The first volume of air inside the tire V_2

= The second volume of air inside the tire = $30\% V_1 = 0.3V_1$ V_3

= The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$ P_1

= The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

Assume:

= The second pressure of air inside the tire P_3

= The third pressure of air inside the tiren = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

Answer from: Quest

The answer is d) by requesting advice on science internet sites

Answer from: Quest

so here is the information for current research:

-nrel's geothermal analysts, scientists, and engineers perform a full range of research from developing advanced methods of locating and accessing geothermal resources to exploring innovative technologies to extract and produce clean energy from the heat of the earth

and here is the information for the technology of geothermal energy:

-it's clean and sustainable

-resources of geothermal energy range from the shallow ground to hot water and hot rock found a few miles beneath the earth's surface, and down even deeper to the extremely high temperatures of molten rock called magma

-almost everywhere, the shallow ground or upper 10 feet of the earth's surface maintains a nearly constant temperature between 50° and 60°f (10° and 16°c)

-geothermal heat pumps can tap into this resource to heat and cool buildings

-a geothermal heat pump system consists of a heat pump, an air delivery system (ductwork), and a heat exchanger-a system of pipes buried in the shallow ground near the building

-in the winter, the heat pump removes heat from the heat exchanger and pumps it into the indoor air delivery system

-in the summer, the process is reversed, and the heat pump moves heat from the indoor air into the heat exchanger

-the heat removed from the indoor air during the summer can also be used to provide a free source of hot water

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