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
mixer [17]
3 years ago
13

When a gas is heated, it absorbs 196 joules of heat from the surroundings. At the same time, the gas expands, doing pressure-vol

ume work (P · V work) on the surroundings that is equal to 322 joules.
(a) Determine whether the amounts of heat (q) and work (w) exchanged should have positive or negative signs. heat (q) positive negative work (w) positive negative

(b) Calculate the change in internal energy (ΔE) of the gas. J

(c) Determine whether one or more of the following is a state function:

internal energy (E) of a system,
change in internal energy (ΔE) of a system,
heat (q) absorbed or released by a system,
work (w) done on or by a system.
Physics
1 answer:
timurjin [86]3 years ago
7 0

Answer:

(a) q positive; w negative.

(b) ΔE = -126 J

(c) E and ΔE

Explanation:

<em>(a) Determine whether the amounts of heat (q) and work (w) exchanged should have positive or negative signs. heat (q) positive negative work (w) positive negative</em>

By convention, when the system absorbs heat from the surroundings, its sign is positive, that is, q = 196 J.

By convention, when the system exerts work on the surroundings, its sign is negative, that is, w = -322 J.

<em>(b) Calculate the change in internal energy (ΔE) of the gas. J</em>

The change in internal energy (ΔE) can be calculated using the following expression.

ΔE = q + w

ΔE = 196 J + (-322 J) = -126 J

<em>(c) Determine whether one or more of the following is a state function: </em>

<em> internal energy (E) of a system, </em>

<em>change in internal energy (ΔE) of a system, </em>

<em>heat (q) absorbed or released by a system, </em>

<em>work (w) done on or by a system.</em>

<em />

E and ΔE are state functions (they only depend on the states of the gas), whereas q and w depend on the trajectory.

You might be interested in
The accepted standard measurement for mass is the A) liter. B) pound. C) Newton. D) kilogram.
andreev551 [17]
D) Kilograms is the standard unit of mass
6 0
3 years ago
An object accelerating may be changing in what two ways
Oksana_A [137]
Velocity and direction
8 0
3 years ago
How are the layers of the atmosphere separated
Volgvan

Answer:

it depends on temperature^^

Explanation:

hope this helped^^

8 0
4 years ago
Read 2 more answers
Earth's inner core is:a dense ball of solid metal
hram777 [196]
What is one effect of steroid abuse in professional sports?
AThey cause athletes to stop training hard incorrect answer
BThey allow some competitors to gain an unfair advantage incorrect answer
CThey cause athletes to stop competing to win incorrect answer
DThey cause athletes to become selfish
7 0
3 years ago
A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.
Nesterboy [21]

Answer:

<em>The velocity after the collision is 2.82 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}

\displaystyle v'=\frac{22.287}{7.91}

\boxed{v' = 2.82\ m/s}

The velocity after the collision is 2.82 m/s

6 0
3 years ago
Other questions:
  • An operation has a 20 percent scrap rate. As a result, 80 pieces per hour are produced. What is the potential labor productivity
    8·1 answer
  • Vertical columns on the periodic table are called
    15·1 answer
  • How does the total momentum of the docked vehicle and station compare to the momentum of each object before the
    13·1 answer
  • A parallel-plate capacitor is charged by connecting it to a battery. After that the capacitor is disconnected from the battery a
    5·1 answer
  • A 12.3 n force is applied to a 3.65 kg object. Determine the acceleration of the object​
    13·1 answer
  • What is the velocity at 11 seconds
    5·1 answer
  • Which would help you perform a basketball skill well with<br> speed?
    14·1 answer
  • Which statement does not describe a characteristic of a good outline?
    15·2 answers
  • Ilang ektarya ang lawak na itinakda ni llda na maaaring gawin palaisdaan​
    5·1 answer
  • What do alcohol, drugs, and tobacco all have in common?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!