Chemistry, 09.12.2019 20:31 carter4026

The temperature of 2.0 mol of a monatomic ideal gas is 340 k. the internal energy of this gas is doubled by the addition of heat. (a) how much heat is needed when it is added at constant volume? j (b) how much heat is needed when it is added at constant pressure?

Answers: 3

Show answers

Answers

Answer from: chitteshchandra56

Answer :

(a) The heat needed when it is added at constant volume is 8.5 kJ.

(b) The heat needed when it is added at constant pressure is 14.1 kJ.

Explanation :

Heat released at constant pressure is known as enthalpy.

Heat released at constant volume is known as internal energy.

(a) The formula used for change in internal energy of the gas is:

$\Delta Q_v=\Delta U\\\\\Delta U=nC_v\Delta T\\\\\Delta Q=nC_v(T_2-T_1)$

where,

$\Delta Q_v$ = heat at constant volume = ?

$\Delta U$ = change in internal energy

n = number of moles of gas = 2.0 moles

C_v = heat capacity at constant volume for monoatomic gas = $\frac{3}{2}R$

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 340 K

T_2 = final temperature = $340\times 2=680K$

Now put all the given values in the above formula, we get:

$\Delta Q_v=nC_v(T_2-T_1)$

$\Delta Q_v=(2.0moles)\times (\frac{3}{2}R)\times (680-340)K$

$\Delta Q_v=(2.0moles)\times (\frac{3}{2}\times 8.314J/mole.K)\times (680-340)K$

$\Delta Q_v=8480.28J=8.5kJ$

Thus, the heat needed when it is added at constant volume is 8.5 kJ.

(b) The formula used for change in enthalpy of the gas is:

$\Delta Q_p=\Delta H\\\\\Delta H=nC_p\Delta T\\\\\Delta Q=nC_p(T_2-T_1)$

where,

$\Delta Q_p$ = heat at constant pressure = ?

$\Delta H$ = change in enthalpy energy

n = number of moles of gas = 2.0 moles

C_p = heat capacity at constant pressure for monoatomic gas = $\frac{5}{2}R$

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 340 K

T_2 = final temperature = $340\times 2=680K$

Now put all the given values in the above formula, we get:

$\Delta Q_p=nC_p(T_2-T_1)$

$\Delta Q_p=(2.0moles)\times (\frac{5}{2}R)\times (680-340)K$

$\Delta Q_p=(2.0moles)\times (\frac{5}{2}\times 8.314J/mole.K)\times (680-340)K$

$\Delta Q_p=14133.8J=14.1kJ$

Thus, the heat needed when it is added at constant pressure is 14.1 kJ.

Answer from: Quest

precipitation

explanation:

when it goes from cloud to rain/back to earth in the form of rain

Answer from: Quest

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

Another question on Chemistry

Chemistry, 21.06.2019 19:30

If the element whose electric configuration ends in the d sublevel, the element is calssified as? a.inner transition b.noble gases c.representative d. transition

Answers: 2

Answer

Chemistry, 22.06.2019 05:30

What is the mass of each element in a 324.8 sample of co2

Answers: 1

Answer

Chemistry, 22.06.2019 19:00

Nan element’s square on the periodic table, the number with the greatest numerical value represents the

Answers: 3

Answer

Chemistry, 23.06.2019 05:00

Scientists discovered fossils in several layers of the earth you see here. they found fossils of algae, snails, and clams in layer d. given that information, where do you think they found fossil evidence of simple land plants and amphibians?

Answers: 1

Answer

You know the right answer?

The temperature of 2.0 mol of a monatomic ideal gas is 340 k. the internal energy of this gas is dou...

Questions