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anzhelika [568]
3 years ago
10

When a fixed amount of ideal gas goes through an isobaric expansion A) its internal (thermal) energy does not change.B) the gas

does no work.C) no heat enters or leaves the gas. D) its temperature must increase.E) its pressure must increase.
Physics
1 answer:
Bingel [31]3 years ago
7 0
<h2>Answer: its temperature must increase.</h2>

Explanation:

In an isobaric process the pressure remains constant, which means the initial pressure and the final pressure will be the same.

In addition, during this thermodynamic process, the volume of the ideal gas expands or contracts in such a way that the variation of pressure \Delta P is neutralized.

Now, according to the First law of Thermodynamics that establishes the conservation of energy:

\Delta U=\Delta Q-\Delta W   (1)

Where:

\Delta U is the internal energy

\Delta Q is the heat transferred

\Delta W is the work

Now, for an isobaric process:

\Delta W=P\Delta V    (2)

Where:

P is the pressure (<u>always positive</u>)

\Delta V is the volume variation of the gas

<u />

<u>Here we have two possible results:</u>

-If the gas expands (positive \Delta V), the work is positive.

-If the gas compresses (negative \Delta V), the work is negative.

In this case we are talking about the first result (work is positive).

Then, according to the above, equation (1) can be written as follows:

\Delta U=\Delta Q - P\Delta V   (3)

Clearing \Delta Q:

\Delta Q=\Delta U+P \Delta V    (4)

Then, for an ideal gas in an isobaric process, part of the heat (Q) added to the system will be used to do work (positive in this case) and the other part <u>will increase the internal energy</u>, hence <u>the temperature will increase as well.</u>

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Based on your observations of the six collisions, describe the physical difference between elastic and inelastic collisions.
Ahat [919]

Answer:

Collisions are basically two types: Elastic, and inelastic collision. Elastic collision is defined as the colliding objects return quickly without undergoing any heat generation. Inelastic collision is defined as the where heat is generated, and colliding objects are distorted.

In elastic collision, the total kinetic energy, momentum are conserved, and there is no wasting of energy occurs. Swinging balls is the good example of elastic collision. In inelastic collision, the energy is not conserved it changes from one form to another for example thermal energy or sound energy. Automobile collision is good example, of inelastic collision.

5 0
2 years ago
A sample of copper has a volume of 23.4 cm3 if the density of copper is 8.9 gcm3 what is the coppers mass?
murzikaleks [220]
The answer is:  " 208 g " .
_____________________________________________
Explanation:
__________________________________________
The formula/ equation for density is:
__________________________________________
D = m / V  ;  That is,  "mass divided by volume" ;
 
Density is expressed as:
__________________________________________    
                   "mass per unit volume";  in which the "mass" is expressed in units of "g" ("grams") ;  and the "unit volume" is expressed in units of:
    "cm³ " or "mL"; 
_____________________________________________
           {Note the exact equivalent:  1 cm³ = 1 mL }.
____________________________________________
         →  The formula is:  " D = m / V "  ; 
___________________________________________
   in which:

     "D" refers to the "density" (see above), which is: "8.9 g/cm³ " (given); 

     "m" refers to the "mass" , in units of "g" (grams), which is unknown; and we want to find this value;
                 
     "V" refers to the "volume", in units of "cm³ " ;
               which is:  "23.4 cm³ " (given);
_________________________________________________
                 We want to find the mass, "m" ; so we take the original equation/formula for the density:
_________________________________________________ 
              D  =  m / V ; 
_________________________________________________________
             And we rearrange; to isolate "m" (mass) on ONE side of the    equation; and then we plug in our known/given values;
 to solve for "m" (mass);  in units of "g" (grams) ;
___________________________________________________
    Multiply each side of the equation by "V" ; 
____________________________________________________
             V * { D  =  m / V } ;  to get:
____________________________________________________
      V * D = m ;   ↔   m = V * D ;
___________________________________________________
           Now, we plug in the given values for "V" (volume) and "D" (density) ;     to solve for the mass, "m" ;
______________________________________________________
           m  =  V * D ;
 
           m  =  (23.4 cm³) * (8.9 g / 1 cm³)  = (23.4 * 8.9) g = 208.26 g ;
  
 →  Round to "208 g" (3 significant figures);  
____________________________________
The answer is:  " 208 g " .
_____________________________________________________
7 0
3 years ago
There are many interesting applications of our energy density model to the flow of blood in the human circulatory system. One in
qaws [65]

Answer:

Pressure increases due to enlargement

Explanation:

Energy density is just a fancy name for pressure

Pressure is same at the bottom of the cups (same level-Pascal's law)

thus, Air pressure 1 + h1d1g = Air pressure 2 + h2d1g

= Air pressure 3 + (h2-h1)d2g +h1d1g

from the first 2, we get that since h2>h1, AP2<AP1

from the next 2, we get that since d2<d1, AP3>AP2

from first and third, we get that AP1>AP3

thus, finally AP1>AP3>AP2

for fluids flowing in tubes (blood vessel in this case)

P+0.5dv^2 + gh is constant (also called the bernoulli equation

for the same blood vessel, the heights remain same i.e h1=h2

for same flow rate, inc in area decreases the speed at which the blood flows as vA must remain same

hence, P increases due to enlargement

5 0
3 years ago
Read 2 more answers
Two people push on the same door from opposite sides as shown.
tangare [24]
The correct answer is
<span>c. one person exerts more force than the other so that the forces are unbalanced.

In fact, the door is initially at rest. In order to move the door, a net force different from zero should be applied, according to Newton's second law:
</span>\sum F = ma
<span>where the term on the left is the resultant of the forces acting on the door, m is the door mass and a its acceleration.

In order to move the door, the acceleration must be different from zero. But this means that the resultant of the forces acting on it must be different from zero: this is possible only if  the forces applied by the two persons are unbalanced, i.e. one person exerts more force than the other.</span>
3 0
3 years ago
Read 2 more answers
A 100 kg box sits on an incline held at rest by a very thin rope attached to another mass hanging as shown. The force of frictio
MAVERICK [17]

If friction is acting along the plane upwards

then in this case we will have

For equilibrium of 100 kg box on inclined plane we have

mgsin\theta = F_f + T

also for other side of hanging mass we have

T = Mg = 50(9.8) = 490 N

now we have

100(9.8)sin\theta = 100 + 490

980sin\theta = 590

sin\theta = 0.602

\theta = 37 degree

In other case we can assume that friction will act along the plane downwards

so now in that case we will have

mgsin\theta + F_f = T

also we have

T = Mg = 50(9.8) N

now we have

100(9.8)sin\theta + 100 = 50(9.8)

980sin\theta + 100 = 490

980 sin\theta = 490 - 100

sin\theta = 0.397

\theta = 23.45 degree

<em>So the range of angle will be 23.45 degree to 37 degree</em>

6 0
3 years ago
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