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As saidstated in other answers, the internal energy $U$ depends on other variables, e.g. volume, if internal interactions are not negligible.

On the other hand, there's a simple experience thatsimple experiment which shows that for nearly ideal gases, such as air in ordinaryat standard conditions, $U$ is a function of only the temperature $T$. (andIt is therefore assumed that for ideal gases, that $U$ is exactly a function of $T$). We)

In the experiment, we have an adiabatic container with a wall that divides it in two comparts,compartments: one side is filled with air at atmospheric pressure, while inand the other is made thea vacuum. NowThen, we remove the wall and let the gas expand. If we measure the temperature of the gas during the process, what we observe is that itthe temperature practically stays constant. Now

Now let's look at the whole process: since. Since the gas is thermally insulated, it doesn't exchange heat with the sourroundingssurroundings: $$Q=0.$$ The work done on the gas during the process is zero, since there's no force that keeps the gas from expanding:$$W=0.$$ So, by the conservation of energy: $$\Delta U = Q+W=0.$$ The net result is a change in the volume occupied by the gas (with an inverse change in pressure), while $\Delta U =0$. So, we conclude that the internal energy is independent fromof the volume, for ideal gases.

As said in other answers, the internal energy $U$ depends on other variables, e.g. volume, if internal interactions are not negligible.

On the other hand, there's a simple experience that shows that for nearly ideal gases, such as air in ordinary conditions, $U$ is a function of only the temperature $T$ (and is therefore assumed that for ideal gases $U$ is exactly a function of $T$). We have an adiabatic container with a wall that divides it in two comparts, one filled with air at atmospheric pressure, while in the other is made the vacuum. Now we remove the wall and let the gas expand. If we measure the temperature of the gas during the process, what we observe is that it practically stays constant. Now let's look at the whole process: since the gas is thermally insulated, it doesn't exchange heat with the sourroundings: $$Q=0.$$ The work done on the gas during the process is zero, since there's no force that keeps the gas from expanding:$$W=0.$$ So, by the conservation of energy: $$\Delta U = Q+W=0.$$ The net result is a change in the volume occupied by the gas (with an inverse change in pressure), while $\Delta U =0$. So we conclude that the internal energy is independent from the volume, for ideal gases.

As stated in other answers, the internal energy $U$ depends on other variables, e.g. volume, if internal interactions are not negligible.

On the other hand, there's a simple experiment which shows that for nearly ideal gases, such as air at standard conditions, $U$ is a function of only temperature $T$. (It is therefore assumed for ideal gases, that $U$ is exactly a function of $T$.)

In the experiment, we have an adiabatic container with a wall that divides it in two compartments: one side is filled with air at atmospheric pressure, and the other is a vacuum. Then, we remove the wall and let the gas expand. If we measure the temperature of the gas during the process, we observe the temperature practically stays constant.

Now let's look at the whole process. Since the gas is thermally insulated, it doesn't exchange heat with the surroundings: $$Q=0.$$ The work done on the gas during the process is zero, since there's no force that keeps the gas from expanding:$$W=0.$$ So, by the conservation of energy: $$\Delta U = Q+W=0.$$ The net result is a change in the volume occupied by the gas (with an inverse change in pressure), while $\Delta U =0$. So, we conclude internal energy is independent of the volume for ideal gases.

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As said in other answers, the internal energy $U$ depends on other variables, e.g. volume, if internal interactions are not negligible.

On the other hand, there's a simple experiencesimple experience that shows that for nearly ideal gases, such as air in ordinary conditions, $U$ is a function of only the temperature $T$ (and is therefore assumed that for ideal gases $U$ is exactly a function of $T$). We have an adiabatic container with a wall that divides it in two comparts, one filled with air at atmospheric pressure, while in the other is made the vacuum. Now we remove the wall and let the gas expand. If we measure the temperature of the gas during the process, what we observe is that it practically stays constant. Now let's look at the whole process: since the gas is thermally insulated, it doesn't exchange heat with the sourroundings: $$Q=0.$$ The work done on the gas during the process is zero, since there's no force that keeps the gas from expanding:$$W=0.$$ So, by the conservation of energy: $$\Delta U = Q+W=0.$$ The net result is a change in the volume occupied by the gas (with an inverse change in pressure), while $\Delta U =0$. So we conclude that the internal energy is independent from the volume, for ideal gases.

As said in other answers, the internal energy $U$ depends on other variables, e.g. volume, if internal interactions are not negligible.

On the other hand, there's a simple experience that shows that for nearly ideal gases, such as air in ordinary conditions, $U$ is a function of only the temperature $T$ (and is therefore assumed that for ideal gases $U$ is exactly a function of $T$). We have an adiabatic container with a wall that divides it in two comparts, one filled with air at atmospheric pressure, while in the other is made the vacuum. Now we remove the wall and let the gas expand. If we measure the temperature of the gas during the process, what we observe is that it practically stays constant. Now let's look at the whole process: since the gas is thermally insulated, it doesn't exchange heat with the sourroundings: $$Q=0.$$ The work done on the gas during the process is zero, since there's no force that keeps the gas from expanding:$$W=0.$$ So, by the conservation of energy: $$\Delta U = Q+W=0.$$ The net result is a change in the volume occupied by the gas (with an inverse change in pressure), while $\Delta U =0$. So we conclude that the internal energy is independent from the volume, for ideal gases.

As said in other answers, the internal energy $U$ depends on other variables, e.g. volume, if internal interactions are not negligible.

On the other hand, there's a simple experience that shows that for nearly ideal gases, such as air in ordinary conditions, $U$ is a function of only the temperature $T$ (and is therefore assumed that for ideal gases $U$ is exactly a function of $T$). We have an adiabatic container with a wall that divides it in two comparts, one filled with air at atmospheric pressure, while in the other is made the vacuum. Now we remove the wall and let the gas expand. If we measure the temperature of the gas during the process, what we observe is that it practically stays constant. Now let's look at the whole process: since the gas is thermally insulated, it doesn't exchange heat with the sourroundings: $$Q=0.$$ The work done on the gas during the process is zero, since there's no force that keeps the gas from expanding:$$W=0.$$ So, by the conservation of energy: $$\Delta U = Q+W=0.$$ The net result is a change in the volume occupied by the gas (with an inverse change in pressure), while $\Delta U =0$. So we conclude that the internal energy is independent from the volume, for ideal gases.

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As said in other answers, the internal energy $U$ depends on other variables, e.g. volume, if internal interactions are not negligible.

On the other hand, there's a simple experience that shows that for nearly ideal gases, such as air in ordinary conditions, $U$ is a function of only the temperature $T$ (and is therefore assumed that for ideal gases $U$ is exactly a function of $T$). We have an adiabatic container with a wall that divides it in two comparts, one filled with air at atmospheric pressure, while in the other is made the vacuum. Now we remove the wall and let the gas expand. If we measure the temperature of the gas during the process, what we observe is that it practically stays constant. Now let's look at the whole process: since the gas is thermally insulated, it doesn't exchange heat with the sourroundings: $$Q=0.$$ The work done on the gas during the process is zero, since there's no force that keeps the gas from expanding:$$W=0.$$ So, by the conservation of energy: $$\Delta U = Q+W=0.$$ The net result is a change in the volume occupied by the gas (with an inverse change in pressure), while $\Delta U =0$. So we conclude that the internal energy is independent from the volume, for ideal gases.