1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
strojnjashka [21]
3 years ago
5

An ideal gas initially at 4.00atm and 350 K is permitted

Physics
1 answer:
Nuetrik [128]3 years ago
6 0

Explanation:

It is given that initially pressure of ideal gas is 4.00 atm and its temperature is 350 K. Let us assume that the final pressure is P_{2} and final temperature is T_{2}.

(a)   We know that for a monoatomic gas, value of \gamma is \frac{5}{3}[/tex].

And, in case of adiabatic process,

                PV^{\gamma} = constant              

also,         PV = nRT

So, here    T_{1} = 350 K,    V_{1} = V,  and   V_{2} = 1.5 V

Hence,      \frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma -1}

         \frac{T_{2}}{350 K} = (\frac{V}{1.5V})^{\frac{5}{3} -1}

          T_{2} = 267 K

Also,   P_{1} = 4.0 atm,   V_{1} = V,  and   V_{2} = 1.5 V

        \frac{P_{2}}{P_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma}

        \frac{P_{2}}{4.0 atm} = (\frac{V}{1.5V})^{\frac{5}{3}}

            P_{2} = 2.04 atm

Hence, for monoatomic gas final pressure is 2.04 atm and final temperature is 267 K.

(b) For diatomic gas, value of \gamma is \frac{7}{5}[/tex].

As,        PV^{\gamma} = constant              

also,         PV = nRT

T_{1} = 350 K,    V_{1} = V,  and   V_{2} = 1.5 V

              \frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma -1}

         \frac{T_{2}}{350 K} = (\frac{V}{1.5V})^{\frac{7}{5} -1}

          T_{2} = 289 K

And,   P_{1} = 4.0 atm,   V_{1} = V,  and   V_{2} = 1.5 V

                \frac{P_{2}}{P_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma}

        \frac{P_{2}}{4.0 atm} = (\frac{V}{1.5V})^{\frac{7}{5}}

            P_{2} = 2.27 atm

Hence, for diatomic gas final pressure is 2.27 atm and final temperature is 289 K.

You might be interested in
A 0.54 kg particle has a speed of 5.0 m/s at point A and kinetic energy of 7.5 J at point B. What is
Wewaii [24]

<u>Answers</u>

(a)  6.75 Joules.

(b)  5.27 m/s

(c) 0.75 Joules


<u>Explanation</u>

Kinetic energy is the energy possessed by a body in motion.

(a) its kinetic energy at A?

K.E = 1/2 mv²

       = 1/2 ×  0.54 × 5²

       = 6.75 Joules.

(b) its speed at point B?

K.E = 1/2 mv²

7.5 = 1/2 × 0.54 × V²

V² = 7.5 ÷ 0.27

     = 27.77778

V = √27.77778

   = 5.27 m/s

(c) the total work done on the particle as it moves from A to B?

Work done = 7.5 - 6.75

                 = 0.75 Joules

4 0
3 years ago
Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o
IrinaVladis [17]

Answer:

The net gravitational force on the mass is 1.27\times 10^{-5}

Explanation:

We have by Newton's law of gravity the force of attraction between masses m_{1},m_{2}

F_{att}=G\frac{m_{1}m_{2}}{r^{2}}

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N

Force of attraction between 435 kg mass and 38 kg mass is

F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N

Thus the net force on mass 38.0 kg is F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}

8 0
4 years ago
A ray of light strikes a mirror at an angle of 38 to the normal. what is the reflected angle
Sveta_85 [38]
The same 38 degree angle just in the opposite direction
6 0
3 years ago
a 100 g cart initially moving at 0.5 m/s collides elastically from a stationary 180 g cart. a) using the equation in the theory,
cupoosta [38]

A 100 g cart is moving at 0.5 m/s that collides elastically from a stationary 180 g cart. Final velocity is calculated to be 0.25m/s.

Collision in which there is no net loss in kinetic energy in the system as a result of the collision is known as elastic collision . Momentum and kinetic energy both are conserved quantities in elastic collisions.

Collision in which part of the kinetic energy is changed to some other form of energy is inelastic collision.

For an elastic collision, we use the formula,

m₁V₁i+ m₂V₂i = m₁V1f + m₂V₂f

For a perfectly elastic collision, the final velocity of the 100g cart will each be 1/2 the velocity of the initial velocity of the moving cart.

Final velocity = 0.5/2

=0.25 m/s.

To know more about elastic collision, refer

brainly.com/question/7694106

#SPJ4

7 0
1 year ago
Needed substance are carried to the body cells by ehat
agasfer [191]
Oh my gosh had the same question
3 0
3 years ago
Other questions:
  • How many milliliters are in 3.65L
    14·2 answers
  • A 3.00 kg object is fastened to a light spring, with the intervening cord passing over a pulley. The pulley is frictionless, and
    15·1 answer
  • A 50 mm diameter steel shaft and a 100 mm long steel cylindrical bushing with an outer diameter of 70 mm have been incorrectly s
    7·1 answer
  • Name the main resistive force that opposes the driving force when a vehicle is moving.
    13·2 answers
  • Energy waves traveling at the speed of light are called _____.
    10·2 answers
  • If the atomic number of an element is 26, then what else is also 26?
    7·2 answers
  • An object has an average acceleration of + 6.24 m/s ^ 2 for 0.300 s . At the end of this time the object's velocity is + 9.31 m/
    9·1 answer
  • How can you make an item that is made of magnetic material become a permanent magnet instead of a temporary magnet?
    7·2 answers
  • (20 points) An engineer is tasked with developing a model to study a cylindrical heat exchanger in a steam system. The prototype
    11·1 answer
  • What size resistor would produce a 3 amp current flow with a 12 volt battery?
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!