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The pressure-volume relationship is an important concept in thermodynamics, especially for understanding the behavior of gases during expansion or compression. In this context, the relationship is described by the equation:\[ p = 5.00 \exp\left[\frac{V_i - V}{a}\right] \]where:- \(p\) is the pressure of the gas,- \(V\) is the volume,- \(V_i\) is the initial volume,- \(a = 1.00\, \mathrm{m}^3\) is a constant.This equation highlights how the pressure decreases exponentially as the volume increases. When the gas expands from the initial volume to the final volume, the pressure changes accordingly. This relationship informs us that as the gas expands, its particles occupy more space, leading to a reduction in pressure. This exponential relationship allows us to calculate the final pressure of the gas after expansion, revealing that it is approximately 1.8395 kPa when the gas has expanded to a volume of 2.00 m³.

The ideal gas law is a fundamental principle in thermodynamics that describes the relationship between pressure, volume, and temperature of a given amount of gas. It is given by the formula:\[ PV = nRT \]where:- \(P\) is the pressure,- \(V\) is the volume,- \(n\) is the number of moles of the gas,- \(R\) is the ideal gas constant (8.314 J/mol·K),- \(T\) is the temperature in Kelvin.Using this law, you can find unknown variables if other variables are known. For this exercise, knowing the final pressure, volume, and number of moles allows us to solve for the final temperature. The conversion from kPa to Pa (since 1 kPa = 1000 Pa) is crucial before applying the ideal gas law. After applying the values, the temperature of the gas is calculated to be approximately 442.49 K.

Work done by a gas during expansion or compression is calculated using the integral of pressure with respect to volume:\[ W = \int_{V_i}^{V_f} p \, dV \]In this specific problem, the expression for pressure as a function of volume is given by:\[ p = 5.00 \exp\left[\frac{V_i - V}{a}\right] \]To find the work done during expansion, you substitute this expression into the equation for work and integrate from the initial to the final volume. This yields:\[ W = 5.00 \left[-\exp(1-V) \right]_{1}^{2} \]Solving this integral gives the work done as approximately 3.1605 kJ. This means energy is transferred by the gas as it expands, moving the particles and exerting a force over the change in volume.

Entropy is a measure of disorder or randomness in a system, and the change in entropy reflects how the distribution of energy in the system has evolved. In this context, the change in entropy \( \Delta S \) can be calculated by considering reversible processes.For an isothermal process where the temperature remains constant while the volume changes, the change in entropy \( \Delta S_T \) is given by:\[ \Delta S_T = nR \ln\left(\frac{V_f}{V_i}\right) \]In this case, using one mole of gas and the R constant, we find the isothermal entropy change is approximately 5.76 J/K.In an isochoric process (constant volume), only the temperature changes, and since this doesn't apply to our final condition (as the volume changes), the entropy change for this segment is effectively zero. Therefore, the total entropy change is mainly from the isothermal expansion
and amounts to around 5.76 J/K.