Ask question

Ask question

All categories

1 day ago

12

Three moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at consta

nt pressure of 6.9 cal/(mol∙K) starts at 300 K and is heated at constant pressure to 320 K, then cooled at constant volume to its original temperature. How much heat flows into the gas during this two-step process?

1 answer:

ValentinkaMS [1.1K]1 day ago

7 0

Answer:

The amount of heat that enters the gas throughout this two-step process totals 120 cal.

Explanation:

Given that,

Moles present = 3

Heat capacity at volume held constant = 4.9 cal/mol.K

Heat capacity at pressure held constant = 6.9 cal/mol.K

Starting temperature = 300 K

Ending temperature = 320 K

We are tasked with determining the heat absorbed by the gas at constant pressure

Employing the heat formula

$\Delta H_{1}=nC_{p}\times\Delta T$

Substituting the values into the equation

$\Delta H_{1}=3\times6.9\times(320-300)$

$\Delta H_{1}=414\ cal$

Next, we calculate the heat absorbed by the gas at constant volume

Using the corresponding heat formula

$\Delta H_{1}=nC_{v}\times\Delta T$

Insert the values into the formula

$\Delta H_{1}=3\times4.9\times(300-320)$

$\Delta H_{1}=-294\ cal$

Now, it's necessary to evaluate the total heat flow into the gas during both steps

Using the total heat formula

$\Delta H_{T}=\Delta H_{1}+\Delta H_{2}$

$\Delta H_{T}=414-294$

$\Delta H_{T}=120\ cal$

Thus, the heat that transfers into the gas throughout this two-step process amounts to 120 cal.

You might be interested in

The amount of energy necessary to remove an electron from an atom is a quantity called the ionization energy, Ei. This energy ca

Keith_Richards [1021]

Answer:

The resulting value is $E_i = 1.5596 *10^{-18} \ J$ .

Explanation:

The question specifies that

The wavelength is $\lambda = 48.2 nm = 48.2 *10^{- 9 }\ m$ .

The velocity is $v = 2.371*10^6 \ m/s$ .

The mass of the electron is $m_e = 9.109*10^{-31} \ kg$ .

The energy of the incoming light is typically depicted mathematically as

$E = \frac{h * c}{\lambda}$ .

Here, c represents the speed of light with the value $c = 3.0 *10^{8} \ m/s$ .

h stands for Planck's constant with a value of $h = 6.62607015 * 10^{-34 } J\cdot s$ .

Thus,

$E = \frac{6.62607015 * 10^{-34 }* 3.0 *10^{8}}{48.2 *10^{- 9 }}$

=> $E = 4.12 *10^{-18} \ J$ .

Typically, kinetic energy is represented as

$E_k = \frac{1}{2} * m_e * v^2$

=> $E_k = \frac{1}{2} * 9.109*10^{-31} * (2.371*10^6 )^2$ .

=> $E_k = 2.56 *0^{-18} \ J$ .

The ionization energy is generally expressed mathematically as

$E_i = 4.12 *10^{-18} - 2.56 *0^{-18}$

=> $E_i = 1.5596 *10^{-18} \ J$ .

8 0

6 days ago

The Hall effect can be used to calculate the charge-carrier number density in a conductor. If a conductor carrying a current of

kicyunya [1011]

Answer:

$6.6*10^{27}e/m^3$

Explanation:

When calculating Hall voltage, it is crucial to have the current, magnetic field strength, length, area, and number of charge carriers available. The Hall voltage can be expressed using the equation:

$V_h = \frac{iB}{neL}$

Where:

i= the current

B= the magnetic field strength

L = the length

n = the number of charge carriers

e= charge of an electron

We need to replace values and solve for n:

$n= \frac{iB}{V_h e L}$

$n= \frac{2*1.2}{4.5*10^{-6}*5^10^{-3}*1.6*10^{-19}}$

$n= 6.6*10^{27}electron.m^{-3}$

As a result, the charge carrier density is $6.6*10^{27}e/m^3$

5 0

7 days ago

A ball is dropped from the top of a cliff. By the time it reaches the ground, all the energy in its gravitational potential ener

ValentinkaMS [1144]

The ball was released from a height of 20 meters

Explanation:

The scenario is as follows:

1. A ball drops from the edge of a cliff.

2. Upon reaching the ground, the energy held in its gravitational potential energy transforms entirely into kinetic energy.

This implies K.E = P.E.

3. The ball impacts the ground at a speed of 20 m/s.

4. The gravitational field strength noted is 10 N/kg.

<pOur goal is to ascertain the height from which the ball was dropped.

<pSince the ball was dropped from a cliff, its initial velocity is 0.<p→ K.E = $\frac{1}{2}m(v^{2}-v_{0}^{2})$

where v is the final velocity, $v_{0}$ is the initial velocity, and m is the mass.

<p→ v = 20 m/s and $v_{0}$ = 0 m/s.<p→ K.E = $\frac{1}{2}m(20^{2}-0^{2})$

→ K.E = $\frac{1}{2}m(400)$

→ K.E = 200 m joules when the ball strikes the ground.

<p→ P.E = mg h

where g is the gravitational field strength, m is mass, and h signifies height.

<p→ g = 10 N/kg.<p→ P.E = m(10)(h)

→ P.E = 10m h joules.

<p→ P.E = K.E.

→ 10m h = 200 m.

Dividing through by 10m yields:

→ h = 20 meters.

The ball was released from a height of 20 meters.

Learn more

To understand more about gravitational potential energy, visit

8 0

4 days ago

Which, if any, of the following statements concerning the work done by a conservative force is NOT true? All of these statements

Yuliya22 [1153]

Answer:

If the starting and ending points are identical, the overall work equals zero.

Explanation:

Option (D) is correct.

A force is considered conservative when the work performed by it while moving an object from point A to point B does not rely on the path taken and remains consistent across all paths. The work done is determined solely by the initial and final locations of the particle. Thus, when the initial and final positions in a conservative field coincide, the work is said to be zero.

8 0

12 days ago

A pole-vaulter is nearly motionless as he clears the bar, set 4.2 m above the ground. he then falls onto a thick pad. the top of

Yuliya22 [1153]

Refer to the diagram below.

Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².

The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s

As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²

Thus, the deceleration magnitude is 82 m/s².

8 0

17 days ago

Read 2 more answers