Chemistry, 10.12.2019 20:31, mandy9386

Asample consisting of 1.00 mol of perfect gas molecules at 27 °c is expanded isothermally from an initial pressure of 3.00 atm to a final pressure of 1.00 atm in two ways: (a) reversibly, and (b) against a constant external pressure of 1.00 atm. evaluate q, w, δu, δh, δs, δssurr, and δstot in each case.

Answers: 3

Show answers

Answers

answered: kimjooin02

(a) For reversible case,

the heat added (q) is 2741.13 J, the work done, w is -2741.13 J, the change in internal energy, ΔU is 0, the enthalpy change, ΔH is 0, the entropy change of the surrounding, $\Delta S_{surr}$ is -9.13 J/K, the entropy change of the system, $\Delta S_{sys}$ is 9.13 J/K, and the total entropy change, $\Delta S_{total}$ is 0.

(b) For external pressure case,

the heat added (q) is 1661.73 J, the work done, w is -1661.73 J, the change in internal energy, ΔU is 0, the enthalpy change, ΔH is 0, the entropy change of the surrounding, $\Delta S_{surr}$ is -5.54 J/K, the entropy change of the system, $\Delta S_{sys}$ is 9.13 J/K , and the total entropy change, $\Delta S_{total}$ is 3.6 J/K.

The given parameters:

Number of moles, n = 1 mol Temperature, T = 27 °C = 27 + 273 = 300 K Initial pressure, P₁ = 3 atm Final pressure P₂ = 1 atm Ideal gas constant, R = 8.314 J/mol.K

(a) The reversible case:

The first law of thermodynamic is given as;

$\Delta U = q + w$

where;

ΔU is change in internal energy

For Isothermal process, ΔU = 0

q = - w

The work done on the gas is calculated as;

$w = - nRT \ ln(\frac{P_i}{P_f} )\\\\w = - (1)(8.314)(300)\ ln(\frac{3}{1} )\\\\w = -2741.13 \ J$

The heat added to the system;

q = -w

q = 2741.13 J

The change in enthalpy is calculated as;

$\Delta H = 0$

The change in entropy is calculated as;

$\Delta S_{sys} = - \Delta S _{surr}\\\\\Delta S_{sys} = nR \ ln(\frac{P_i}{P_f} )\\\\\Delta S_{sys} = (1)(8.314)\ ln(\frac{3}{1} )\\\\\Delta S_{sys} = 9.13 \ J/K\\\\ \Delta S _{surr} = -9.13 \ J/K$

$\Delta S _{total}= \Delta S _{sys}+ \Delta S _{surr}\\\\\Delta S _{total}= 9.13 - 9.13\\\\\Delta S _{total}= 0$

(b) For a constant external pressure:

$w = -P \Delta V\\\\w = - P (V_f - V_i)$

The initial volume of the gas is calculated as;

$P_iV_i = nRT\\\\V_i = \frac{nRT}{P_i} \\\\V_i = \frac{1 \times 8.314 \times 300}{3 atm\times (101325 \ Pa/atm)} \\\\V_i = 0.0082 \ m^3$

The final volume of the gas is calculated as;

$V_f = \frac{1 \times 8.314 \times 300}{1 atm\times (101325 \ Pa/atm)} \\\\V_f = 0.0246 \ m^3$

The work done is calculated as;

$w =- P(V_f - V_i)\\\\w = -101325 (0.0246 - 0.0082)\\\\w = -1661.73 \ J$

The change in internal energy;

$\Delta U = 0$

The heat added to the system;

q = - w

q = 1661.73 J

The total entropy change is calculated as;

$\Delta S_{surr} = \frac{-q}{T} \\\\\Delta S_{surr} =\frac{-1661.73}{300} \\\\\Delta S_{surr} =-5.54 \ J$

$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} \\\\\Delta S_{total} = 9.13 - 5.54\\\\\Delta S_{total} = 3.6 \ J/K$

Learn more here:link

answered: hxmiltrxshh

a) reversibly

ΔU = 0

q = 2740.16 J

w = -2740.16 J

ΔH = 0

ΔS(total) = 0

ΔS(sys) =9.13 J/K

ΔS(surr) = -9.13 J/K

b) against a constant external pressure of 1.00 atm

ΔU = 0

w = -1.66 kJ

q = 1.66 kJ

ΔH = 0

ΔS(sys) = 9.13 J/K

ΔS(surr) = -5.543 J/K

ΔS(total) = 3.587 J/K

Explanation:

Step 1: Data given:

Number of moles = 1.00 mol

Temperature = 27.00 °C = 300 Kelvin

Initial pressure = 3.00 atm

Final pressure = 1.00 atm

The gas constant = 8.31 J/mol*K

(a) reversibly

Step 2: Calculate work done

For ideal gases ΔU depends only on temperature. So as it is an isothermal (T constant).

Since the temperature remains constant:

ΔU = 0

ΔU = q + w

q = -w

w = -nRT ln (Pi/Pf)

⇒ with n = the number of moles of perfect gas = 1.00 mol

⇒ with R = the gas constant = 8.314 J/mol*K

⇒ with T = the temperature = 300 Kelvin

⇒ with Pi = the initial pressure = 3.00 atm

⇒ with Pf = the final pressure = 1.00 atm

w =- 1*8.314 *300 * ln(3)

w = -2740.16 J

q = -w

q = 2740.16 J

Step 3: Calculate change in enthalpy

Since there is no change in energy, ΔH = 0

Step 4: Calculate ΔS

for an isothermal process

ΔS (total) = ΔS(sys) + ΔS(surr)

ΔS(sys) = -ΔS(surr)

ΔS(sys) = n*R*ln(pi/pf)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -9.13 J/K

ΔS (total) = ΔS(sys) + ΔS(surr) = 0

(b) against a constant external pressure of 1.00 atm

Step 1: Calculate the work done

w = -Pext*ΔV

w = -Pext*(Vf - Vi)

⇒ with Vf = the final volume

⇒ with Vi = the initial volume

We have to calculate the final and initial volume. We do this via the ideal gas law P*V=n*R*T

V = (n*R*T)/P

Initial volume = (n*R*T)/Pi

⇒ Vi = (1*0.08206 *300)/3

⇒ Vi = 8.206 L

Final volume = (n*R*T)/Pf

⇒ Vf = (1*0.08206 *300)/1

⇒ Vf = 24.618 L

The work done w = -Pext*(Vf - Vi)

w = -1.00* ( 24.618 - 8.206)

w = -16.412 atm*L

w = -16 .412 *(101325/1atm*L) *(1kJ/1000J)

w = -1662.9 J = -1.66 kJ

Step 2: Calculate the change in internal energy

ΔU = 0

q = -w

q = 1.66 kJ

ΔH = 0 because there is no change in energy

Step 3: Calculate ΔS

ΔS(sys) = n*R*ln(3)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -q/T

ΔS(surr) = -1662.9J/300K

ΔS(surr) = -5.543 J/K

ΔS(total) = ΔS(surr) +ΔS(sys) = -5.543 J/K + 9.13 J/K = 3.587 J/K

answered: Guest

c. 75.7

i think that is the right

answered: Guest

Elias is almost correct. he just got it backwards.the correct answer is that it demonstrates elastic potential energy transforming into kinetic energy.