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Blue Various
Blue Various
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Another similar stack assumes the absence of the mass and the friction of the piston, but makes no specification regarding the change in temperature in the syringe. Here, we specify the temperature inside the syringe; the gas in the syringe is assumed to be isothermally reversible.

The situations to be considered are as follows;

1 mol of ideal gas is isothermally expanded from $5.05 \times 10^5$ Pa to $1.01 \times 10^5$ Pa at 20 °C against a constant external pressure of $1.01 \times 10^5$ Pa.

My question is as follows;

  • Is the above situation　physically possible ?
  • If so, what controls could be used to achieve this change?

In this case,
the work that the gas in the syringe receives from the outside is; $$W_{outside \to inside} =- \int_{V_1}^{V_2} P_{out} dV = -P_{out}(V_2 - V_1) $$ On the other hand, if this can be regarded as an isothermal reversible process the work received by the outside from the gas inside the syringe is; $$W_{inside \to outside} =- \int_{V_1}^{V_2} P_{in} dV = -nRT\ln({V_2}/{V_1}) $$

This means that there must be a reasonable place for the energy equivalent to the difference between $W_{outside \to inside}$ and $W_{inside \to outside}$ to go.

Another similar stack assumes the absence of the mass and the friction of the piston, but makes no specification regarding the change in temperature in the syringe. Here, we specify the temperature inside the syringe; the gas in the syringe is assumed to be isothermally reversible.

The situations to be considered are as follows;

1 mol of ideal gas is isothermally expanded from $5.05 \times 10^5$ Pa to $1.01 \times 10^5$ Pa at 20 °C against a constant external pressure of $1.01 \times 10^5$ Pa.

My question is as follows;

  • Is the above situation　physically possible ?
  • If so, what controls could be used to achieve this change?

In this case,
the work that the gas in the syringe receives from the outside is; $$W_{outside \to inside} =- \int_{V_1}^{V_2} P_{out} dV = -P_{out}(V_2 - V_1) $$ On the other hand, the work received by the outside from the gas inside the syringe is; $$W_{inside \to outside} =- \int_{V_1}^{V_2} P_{in} dV = -nRT\ln({V_2}/{V_1}) $$

This means that there must be a reasonable place for the energy equivalent to the difference between $W_{outside \to inside}$ and $W_{inside \to outside}$ to go.

Another similar stack assumes the absence of the mass and the friction of the piston, but makes no specification regarding the change in temperature in the syringe. Here, we specify the temperature inside the syringe; the gas in the syringe is assumed to be isothermally.

The situations to be considered are as follows;

1 mol of ideal gas is isothermally expanded from $5.05 \times 10^5$ Pa to $1.01 \times 10^5$ Pa at 20 °C against a constant external pressure of $1.01 \times 10^5$ Pa.

My question is as follows;

  • Is the above situation　physically possible ?
  • If so, what controls could be used to achieve this change?

In this case,
the work that the gas in the syringe receives from the outside is; $$W_{outside \to inside} =- \int_{V_1}^{V_2} P_{out} dV = -P_{out}(V_2 - V_1) $$ On the other hand, if this can be regarded as an isothermal reversible process the work received by the outside from the gas inside the syringe is; $$W_{inside \to outside} =- \int_{V_1}^{V_2} P_{in} dV = -nRT\ln({V_2}/{V_1}) $$

This means that there must be a reasonable place for the energy equivalent to the difference between $W_{outside \to inside}$ and $W_{inside \to outside}$ to go.

Source Link
Blue Various
Blue Various
  • 290
  • 1
  • 10

Work done by the piston versus work done by the surrounding, but, the gas in the piston changes isothermally

Another similar stack assumes the absence of the mass and the friction of the piston, but makes no specification regarding the change in temperature in the syringe. Here, we specify the temperature inside the syringe; the gas in the syringe is assumed to be isothermally reversible.

The situations to be considered are as follows;

1 mol of ideal gas is isothermally expanded from $5.05 \times 10^5$ Pa to $1.01 \times 10^5$ Pa at 20 °C against a constant external pressure of $1.01 \times 10^5$ Pa.

My question is as follows;

  • Is the above situation　physically possible ?
  • If so, what controls could be used to achieve this change?

In this case,
the work that the gas in the syringe receives from the outside is; $$W_{outside \to inside} =- \int_{V_1}^{V_2} P_{out} dV = -P_{out}(V_2 - V_1) $$ On the other hand, the work received by the outside from the gas inside the syringe is; $$W_{inside \to outside} =- \int_{V_1}^{V_2} P_{in} dV = -nRT\ln({V_2}/{V_1}) $$

This means that there must be a reasonable place for the energy equivalent to the difference between $W_{outside \to inside}$ and $W_{inside \to outside}$ to go.