PHYSICS COLLEGE

We start with 5.00 moles of an ideal monatomic gas with an initial temperature of 129 ∘C. The gas expands and, in the process, absorbs an amount of heat equal to 1180 J and does an amount of work equal to 2160 J . What is the final temperature Tfinal of the gas? Use R = 8.3145 J/(mol⋅K) for the ideal gas constant.

Answers

Answer 1

Answer:

The final temperature of an ideal monatomic gas with an initial temperature of 128°C. is 114.53°C.

From the first law of thermodynamics,

ΔU=Q - W

Where,

ΔU - change in internal energy

Q - energy absorbed

W - work

So,

ΔU = 1180 J - 2020 J

ΔU = -840 J

From ideal gas law

$\bold {\Delta U = \frac 32n R (T_2- T_1)}}\n\n\bold {T_ 2 = \frac {2\Delta U}{3nR} +T_1}$

Where, T2 is the final temperature,

n- moles of gas

R - gas constant

T1 - initial temperature,

Put the values in the equation

$\bold {T_ 2 = \frac {2* -840\ J )}{3* 5 * 8.314\ J/mol.K} + 128^oC}\n\n\bold {T_2 = 114.53 ^oC}$

Therefore, the final temperature of an ideal monatomic gas with an initial temperature of 128°C. is 114.53°C.

To know more about ideal gas law,

brainly.com/question/6534096

Answer 2

Answer:

The solution is in the attachment

Related Questions

Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was ?s = 1. Around 1962, three companies independently developed racing tires with coefficients of 1.6. This problem shows that tires have improved further since then. The shortest time interval in which a piston-engine car initially at rest has covered a distance of one-quarter mile is about 4.43 s. (A) Assume the car's rear wheels lift the front wheels off the pavement as shown in the figure above. What minimum value of ?s is necessary to achieve the record time?
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An infant throws 7 g of applesauce at a velocity of 0.5 m/s. All of the applesauce collides with a nearby wall and sticks to it. What is the decrease in kinetic energy of the applesauce?
If 80 joules of work were necessary to move a 5 newton box, how far was the box moved?
Two non-conducting slabs of infinite area are given a charge-per-unit area of σ = -16 C/m^2 and σb =+6.0 C/m^2 respectively. A third slab, made of metal, is placed between the first two plates. The charge density σm on the metal slab is 0 (i.e., the slab is uncharged).Required:Find the magnitude and direction of the electric field.

How would a spinning disk's kinetic energy change if its moment of inertia was five times larger but its angular speed was five times smaller

Answers

A spinning disk's kinetic energy will change to one-tenth if its moment of inertia was five times larger but its angular speed was five times smaller.

Relation between Kinetic energy and Moment of Inertia:

Rotational kinetic energy is directly proportional to the rotational inertia and the square of the magnitude of the angular velocity.

Now, let's consider moment of inertia = I and angular speed = ω

It is asked that what would be change in Kinetic energy if

moment of inertia = (five times larger)

angular speed = ω/5 (five times smaller)

The kinetic energy of a spinning body is given as:

$K.E.=(1)/(2) I. w^2$

On substituting the values, we will get:

$K.E.= (1)/(2) (5I) ((w)/(5) )^2 \n\nK.E. =(1)/(10) I. w^2$

Kinetic energy will be one-tenth to the kinetic energy before its spinning characteristics were changed.

Learn more:

brainly.com/question/12337396

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the

Answers

Complete question:

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

$T_2 = T_1((P_2)/(P_1))^{(k-1 )/(k)$

k = 1.4

$T_2 = T_1((P_2)/(P_1))^{(k-1 )/(k)}\n\nT_2 = 1200((80)/(150))^{(1.4-1 )/(1.4)}\n\nT_2 = 1002.714K$

Work done is given as;

$W = (1)/(2) *m*(v_i^2 - v_e^2)$

inlet velocity is negligible;

$v_e = \sqrt{(2W)/(m) } = √(2*C_p(T_1-T_2)) \n\nv_e = √(2*1004(1200-1002.714))\n\nv_e = √(396150.288) \n\nv_e = 629.41 \ m/s$

Therefore, the exit velocity is 629.41 m/s

As a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 2.43 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 24 m/s. One horsepower equals 746 W.

Answers

Answer:

F=75.53N

Explanation:

To calculate the power we define the equation,

$P=Fv$

Where,

F= Force

V= Velocity,

Here we have that 2.43hp is equal to 1812.78W,

clearing F,

$F=(P)/(v) = (1812)/(24)\nF=75.53N$

A galilean telescope adjusted for a relaxed eye is 36cm long. If the objective lens has a focal length of 40cm, what is the magnification?

Answers

For this problem, we use the mirror equation which is expressed as:

1/di + 1/f = 1/d0

Magnification is expressed as the ratio of di and d0.

Manipulating the equation, we will have:

M = di/f +1
M = 36/40 + 1
M = 1.9

Hope this answers the equation.

A uniform 190 g rod with length 43 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 38 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 10 cm on each sides of the center, at which time the system rotates at an angular speed of 12 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.

Answers

Answer:

The angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s

Explanation:

Moment of inertia is given as;

I = ¹/₁₂×ML² + 2mr²

where;

I is the moment of inertia

M is the mass of the rod = 0.19 kg

L is the length of the rod = 0.43 m

m is the mass of the bead = 0.038 kg

r is the distance of one bead

Initial moment of inertial is given as;

$I_i = (1)/(12)ML^2 +2mr_1^2$

Final moment of inertia is also given as

$I_f= (1)/(12)ML^2 +2mr_2^2$

Angular momentum is the product of angular speed and moment of inertia;

= Iω

From the principle of conservation of angular momentum;

$I_i \omega_i = I__f } \omega_f$

$((1)/(12)ML^2 +2mr_1^2) \omega_i = ((1)/(12)ML^2 +2mr_2^2) \omega_f$

Given;

ωi = 12 rad/s

r₁ = 10.0 cm = 0.1 m

r₂ = 10.0cm/4 = 2.5 cm = 0.025 m

Substitute these values in the above equation, we will have;

$((1)/(12)*0.19*(0.43)^2 +2*0.038(0.1)^2) 12 = ((1)/(12)*0.19*(0.43)^2 +2*0.038*(0.025)^2) \omega_f\n\n0.04425 =0.002975\ \omega_f\n\n\omega_f = (0.04425)/(0.002975) = 14.87\ rad/s$

Therefore, the angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s

Which is true about inelastic collisions: a. An inelastic collision does not obey conservation of energy. b. An inelastic collision conserves kinetic energy. c. Objects will stick together upon collision. d. Momentum is not conserved in inelastic collisions..

Answers

Answer:

Option c is correct

Explanation:

There are two types of collisions-elastic collision and inelastic collision.

In elastic collision, both kinetic energy and total momentum are conserved. On the other hand, in inelastic collision, total momentum is conserved but kinetic energy is not conserved. Thus, option b and d are incorrect.

Total energy is always conserved in both types. Thus, option a is incorrect.

In a perfectly inelastic collision, objects stick together. This happens because maximum kinetic energy is dissipated and used in bonding of the two objects. Thus, correct option is c.

Answer:

i believe its a?

Explanation:

In an inelastic collision, momentum is conserved

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