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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.

Respuesta :

danialamin

Answer:

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

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

[tex]P_{N_2}=P_{O_2}[/tex]

As the ideal gas equation is valid so

[tex]P_{N_2}=\frac{n_{N_2}RT_{N_2}}{V_{{N_2}_f}}[/tex]

Here

n is the moles of [tex]N_2[/tex] 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.

[tex]P_{O_2}=\frac{n_{O_2}RT_{O_2}}{V_{{O_2}_f}}[/tex]

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

[tex]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}}[/tex]

Also as the total volume will remain the same so

[tex]{V_{{N_2}_f}}+{V_{{O_2}_f}}=2.5 \times 10^{-3} m^3[/tex]

Using the value of nitrogen volume in the above equation leads

[tex]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[/tex]

Solving for nitrogen gas volume

[tex]{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[/tex]

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

Part b

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

[tex]W_{N_2}=n_{N_2}RTln\frac{V_{N_2_f}}{V_{N_2_i}}[/tex]

Here

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

So solving the equation gives

[tex]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[/tex]

Similarly for Oxygen

[tex]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[/tex]

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

[tex]\Delta U=\frac{f}{2} nR\Delta T[/tex]

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

[tex]\Delta Q=W+\Delta U[/tex]

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

For nitrogen gas

[tex]\Delta Q_{N_2}=W_{N_2}\\[/tex]

Substituting the value of work done as calculated in part b

[tex]\Delta Q_{N_2}=W_{N_2}\\\Delta Q_{N_2}=-11.13 J\\[/tex]

For oxygen gas

[tex]\Delta Q_{O_2}=W_{O_2}\\[/tex]

Substituting the value of work done as calculated in part b

[tex]\Delta Q_{O_2}=W_{O_2}\\\Delta Q_{O_2}=13.64 J\\[/tex]

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