Physics, 15.02.2020 04:00 lizzytyler4725

52. A metallic container of fixed volume of 2.5 × 10−3 m3 immersed in a large tank of temperature 27 °C contains two compartments separated by a freely movable wall. Initially, the wall is kept in place by a stopper so that there are 0.02 mol of the nitrogen gas on one side and 0.03 mol of the oxygen gas on the other side, each occupying half the volume. When the stopper is removed, the wall moves and comes to a final position. The movement of the wall is controlled so that the wall moves in infinitesimal quasi-static steps. (a) Find the final volumes of the two sides assuming the ideal gas behavior for the two gases. (b) How much work does each gas do on the other? (c) What is the change in the internal energy of each gas? (d) Find the amount of heat that enters or leaves each gas.

Answers: 3

Show answers

Answers

Answer from: Jasten

Part a : The final volume of oxygen gas is $1.5 \times 10^{-3} m^3$ whereas that of nitrogen is $1.0 \times 10^{-3} m^3$

Part b: The work done by oxygen and nitrogen is 13.64 J and -11.13 J respectively.

Part c: The change of internal energy in this process is zero.

Part d: 13.62 J of heat enters, oxygen and 11.11 J of heat leaves Nitrogen.

Explanation:

Part a

As gases behave ideally and the process is taken in the quasi static steps thus at equilibrium

$P_{N_2}=P_{O_2}$

As the ideal gas equation is valid so

$P_{N_2}=\frac{n_{N_2}RT_{N_2}}{V_{{N_2}_f}}$

Here

n is the moles of N_2 which is given as 0.02 mol.

R is the general gas constant which is 8.3

T is the temperature of the gas which is given a 27C or 300K

V is the volume of nitrogen gas at equilibrium which is to be calculated.

$P_{O_2}=\frac{n_{O_2}RT_{O_2}}{V_{{O_2}_f}}$

Here

n is the moles of oxygen which is given as 0.03 mol.

R is the general gas constant which is 8.3

T is the temperature of the gas which is given a 27C or 300K

V is the volume of oxygen gas at equilibrium which is to be calculated.

As both the pressures would be equal so

$P_{O_2}=P_{N_2}\\\frac{n_{O_2}RT_{O_2}}{V_{{O_2}_f}}=\frac{n_{N_2}RT_{N_2}}{V_{{N_2}_f}}\\\frac{n_{O_2}}{V_{{O_2}_f}}=\frac{n_{N_2}}{V_{{N_2}_f}}\\{V_{{N_2}_f}}=0.67{V_{{O_2}_f}}$

Also as the total volume will remain the same so

${V_{{N_2}_f}}+{V_{{O_2}_f}}=2.5 \times 10^{-3} m^3$

Using the value of nitrogen volume in the above equation leads

$0.67{V_{{o_2}_f}}+{V_{{O_2}_f}}=2.5 \times 10^{-3} m^3\\1.67{V_{{o_2}_f}}=2.5 \times 10^{-3} m^3\\{V_{{o_2}_f}}=1.5 \times 10^{-3} m^3$

Solving for nitrogen gas volume

${V_{{N_2}_f}}=0.67{V_{{O_2}_f}}=0.67 \times 1.5 \times 10^{-3} m^3\\{V_{{N_2}_f}}=1.0 \times 10^{-3} m^3$

The final volume of oxygen gas is $1.5 \times 10^{-3} m^3$ whereas that of nitrogen is $1.0 \times 10^{-3} m^3$

Part b

As the process is with constant temperature, the work by nitrogen is given as

$W_{N_2}=n_{N_2}RTln\frac{V_{N_2_f}}{V_{N_2_i}}$

Here

$V_{N_2_i}$ is half of the total volume i.e $1.25 \times 10^{-3} m^3$

So solving the equation gives

$W_{N_2}=0.02\times 8.31 \times 300 ln\frac{1.0\times 10^{-3}}{1.25\times 10^{-3}}\\W_{N_2}=-11.13 J$

Similarly for Oxygen

$W_{O_2}=n_{O_2}RTln\frac{V_{O_2_f}}{V_{O_2_i}}\\W_{O_2}=0.03\times 8.31 \times 300 ln\frac{1.5\times 10^{-3}}{1.25\times 10^{-3}}\\W_{O_2}=13.64 J$

The work done by oxygen and nitrogen is 13.64 J and -11.13 J respectively.

Part c

As the change in internal energy is given as

$\Delta U=\frac{f}{2} nR\Delta T$

In this process as the temperature is constant throughout i.e.T=300K so ΔT=0

Thus the change of internal energy is 0.

The change of internal energy in this process is zero.

Part d

As Heat is given as

$\Delta Q=W+\Delta U$

As ΔU is zero as calculated above in part c so than the heat is given as

For nitrogen gas

$\Delta Q_{N_2}=W_{N_2}\\$

Substituting the value of work done as calculated in part b

$\Delta Q_{N_2}=W_{N_2}\\\Delta Q_{N_2}=-11.13 J\\$

For oxygen gas

$\Delta Q_{O_2}=W_{O_2}\\$

Substituting the value of work done as calculated in part b

$\Delta Q_{O_2}=W_{O_2}\\\Delta Q_{O_2}=13.64 J\\$

13.62 J of heat enters, oxygen and 11.11 J of heat leaves Nitrogen

Answer from: Quest

3.2x10^7

Answer from: Quest

i think it would be 5

Another question on Physics

Physics, 22.06.2019 05:30

Choose the most likely outcome of this scenario: jen decided to go bike riding without a helmet. while no one is around during her ride, she is thrown from her bike when her wheel goes into a pothole. she is not injured, but she is terrified to get back on her bike. what happens next? a. her physical health is affected even though she wasn't hurt. b. her mental and emotional health are affected because she is afraid to get back on her bike. c. her social health is affected because she is worried her friends saw the fall. d. her overall health is not affected at all by her fall.

Answers: 1

Answer

Physics, 22.06.2019 11:30

While you are driving in the lane next to the curb on a multi-lane road the car on your left suddenly moves toward you lane. they are about toy crash into your front fender. you

Answers: 2

Answer

Physics, 22.06.2019 13:00

Nacidified solution was electrolyzed using copper electrodes. a constant current of 1.18 a caused the anode to lose 0.584 g after 1.52 ✕ 103 s. given that the charge of an electron is 1.6022 ✕ 10−19 c, calculate avogadro's number. assume that copper is oxidized to cu2+ ions.

Answers: 1

Answer

Physics, 22.06.2019 19:30

The ability to make things happen is also called a. heat b. force c. matter d. energy

Answers: 1

Answer

You know the right answer?

52. A metallic container of fixed volume of 2.5 × 10−3 m3 immersed in a large tank of temperature 27...

Questions