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1 year ago

How do I do this partial pressure problem? (AP CHEM)

Chemistry

1 answer:

RoseWind [281]1 year ago

7 0

The partial pressure of H(g) is 1.07 atm.

<h3>What is the partial pressure?</h3>

We know that the total pressure of a mixture of gases is the sum of the individual pressure of the gases. Now we know from the Dalton law of partial pressure that the total pressure PT = PA + PB + PC + ........

Thus, the molecular hydrogen has a pressure of 2.14 atm and this molecular hydrogen is made to decompose. The partial pressure of each of the H(g) is 2.14/2 = 1.07 atm.

Hence, the partial pressure of H(g) is 1.07 atm.

Learn more about partial pressure:brainly.com/question/13199169

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The disturbance of a supersaturated solution will cause

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Nicotine is a toxic substance present in tobacco leaves. There are two lone pairs in the structure of nicotine. In general, loca

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3 years ago

A sample consisting of 1.00 mol of perfect gas molecules at 27 °C is expanded isothermally from an initial pressure of 3.00 atm

Evgesh-ka [11]

Answer:

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

4 0

3 years ago