Four per day in an isil thermal process involving an ideal gas. The work done on the gas w would be equal in the negative work in the environment. This would be equaling negative and are tee times the natural log of the final volume divided by the initial volume. And this is again I so thermal process with an ideal gas from this, when the temperature is constant, however, the ideal gas law is then pressure initial volume, initial equaling pressure, final volume, final and this is equaling anarchy when the temperature is constant and so we we we can rewrite the work done on the gas. This would be equal in the negative initial pressure initial volume multiplied by the natural log of the final volume divided by the initial volume. So this would be equaling negative 10 to the third. Pascal's multiplied by 0.500 square rather cubic meters, multiplied by the natural log of 1.25 cubic meters, divided by again 0.500 cubic meters, and we find that the work done is then equaling negative 4.58 times 10 to the fourth jewels. The fact that this is negative means that the gas is doing work on the environment. No, for part B, um, we have simply the first law of thermodynamics with a change in the internal energy is equaling the work minus rather the work, plus the heat transfer, plus the work, My apologies. And so they heat the heat transfer in this thermal isil thermal expansion. We could say Q would be equaling the change in the internal energy minus the work done. And here it's isil thermal expansion. So the property of a nice oh, thermal expansion, any isil thermal process. The change in the internal energy is gonna be zero. And so this would be equaling negative negative four point 58 times 10 to the fourth Jules. And this is giving us 4.58 times turn to the fourth jewels. So positive. So he here work is being done on the environment. Therefore, work is being done on the environment. Therefore, heat is being he is entering the system. Thermal energy is being added to the system. Uh, given that again, this is a nice so thermal process and for part C ah, we can immediately know again the change in the internal energy is zero Jules, and this is again for any I so thermal process. That is the end of the solution. Thank you for watching.