4. Sebanyak 1,05 mol gas ideal, berubah keadaannya secara isoterm dari V_(1)=23 liter dan P_(1)=1,02 atm menjadi V_(2)=24 liter dan P_(2)=0,9775atm. . Hitung Delta U , W, dan Q dalam proses ini

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Pertanyaan

4. Sebanyak 1,05 mol gas ideal, berubah keadaannya secara isoterm dari V_(1)=23 liter dan P_(1)=1,02 atm menjadi V_(2)=24 liter dan P_(2)=0,9775atm. . Hitung Delta U , W, dan Q dalam proses ini

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Answer 1

1. ΔU = 0 J (since the process is isotermic)
2. W = nRT ln(V2/V1) = 1.05 mol * 8.314 J/(mol*K) * 273.15 K * ln(24 L / 23 L) = 1.05 J
3. Q = ΔU + W = 0 J + 1.05 J = 1.05 J

Answer 2

This is a question about the first law of thermodynamics applied to an ideal gas. The first law of thermodynamics states that the change in internal energy of a system (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W). In this case, we are given the initial and final volumes (V1 and V2) and pressures (P1 and P2) of the gas, as well as the amount of gas in moles (n). We can use these values to calculate ΔU, W, and Q.

1. Calculate ΔU: For an ideal gas, the internal energy (U) is directly proportional to the temperature (T). Since the process is isotermic, the temperature does not change, and therefore ΔU is zero.

2. Calculate W: The work done by the gas is given by the formula W = PΔV, where P is the pressure and ΔV is the change in volume. We can calculate ΔV as V2 - V1. The pressure P can be calculated using the ideal gas law PV = nRT, where R is the ideal gas constant. We can rearrange this formula to find P = nRT / V.

3. Calculate Q: The heat added to the system is equal to the change in internal energy, which we have already determined to be zero. Therefore, Q is also zero.

Let's calculate these values using the given data and the formulas above.

Pertanyaan

4. Sebanyak 1,05 mol gas ideal, berubah keadaannya secara isoterm dari V_(1)=23 liter dan P_(1)=1,02 atm menjadi V_(2)=24 liter dan P_(2)=0,9775atm. . Hitung Delta U , W, dan Q dalam proses ini

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