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A 2.35 mole sample of an ideal gas, for which $C_{\mathrm m,v}=3R/2$ initially at $27\ \mathrm{^\circ C}$ and $1750\ \mathrm{kPa}$, undergoes a two stage transformation. For each of the stages described in the following list, calculate the final pressure as well as $q$, $w$, $\Delta U$ and $\Delta H$. Also calculate $q$, $w$, $\Delta U$ and $\Delta H$ for the entire process.

(a) The gas is expanded isothermally and reversibly until the volume triples.
(b) Beginning at the end of the first stage, the temperature is raised to $105\ \mathrm{^\circ C}$ at constant volume.

What I've tried:

Since the process is isothermal, $ΔH=0$. In order to calculate the final pressure. I uses the formula, $P_{\mathrm i}V_{\mathrm i}^{x}=P_{\mathrm f}V_{\mathrm f}^{x}$, where $x=C_{\mathrm m,p}/C_{\mathrm m,v}$. Hence $x=5/3$. I then subsitute this into the formula to get

$$P_{\mathrm i}V_{\mathrm i}^{5/3}=P_{\mathrm f}V_{\mathrm f}^{5/3}$$. Since the volume triples, $V_{\mathrm f}/V_{\mathrm i}=3$ hence manipulating the formula I got $$P_{\mathrm i}/P_{\mathrm f}=3^{5/3}$$ and since $P_{\mathrm i}=1750\ \mathrm{kPa}$. I subsitute this into the above equation to get $P_{\mathrm f}$, which answers the first part of the question.

However I don't seem to get the right answer, hence I could not continue doing the question. Could anyone explain?

Also Am I right to say that we use $P_{\mathrm i}V_{\mathrm i}^{x}=P_{\mathrm f}V_{\mathrm f}^{x}$ for an reversible adiabatic process only. $w=-nRT\ln{\left(T_{\mathrm f}/T_{\mathrm i}\right)}$ for an isothermal process and $\Delta U=nC_{\mathrm m,v}(T_{\mathrm f}-T_{\mathrm i})=-P_\text{external}(V_{\mathrm f}-V_{\mathrm i})$ for a non reversible adiabatic process.

Im rather confused as to decide under which conditions should I use each of these three formulas.

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Hint: In a) the gas is expanding isothermally ($\Delta T=0$). You're computing it as though it's expanding adiabatically ($q=0$). The two processes are quite different! You can use Boyle's law to compute the final pressure for a), and then Amonton's law $P_1/T_1 = P_2/T_2$ to compute the final pressure after process b).

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  • $\begingroup$ So can i use that work formula, $w=-nRTln(V_{f}/V_{i})$ $\endgroup$ – wei jit Jan 26 '15 at 15:13
  • $\begingroup$ Yes, that applies to isothermal ideal gas processes (process a) only). $\endgroup$ – Fred Senese Jan 26 '15 at 15:15
  • $\begingroup$ @weijit Your computation for adiabatic expansion is correct btw. Note however, as Fred (is it okay to say Fred?) posted in his answer that for adiabatic expansion $q=0$, not $\Delta H = 0$. +1 $\endgroup$ – Jori Jan 26 '15 at 15:18
  • $\begingroup$ @Jon, Sure. I've been called worse. $\endgroup$ – Fred Senese Jan 26 '15 at 15:22
  • $\begingroup$ Am i right to say that we use $P_{i}V_{i}^{x}=P_{f}V_{f}^{x}$ for an reversible adiabatic process only. $w=-nRTln(T_{f}/T_{i})$ for an isothermal process and $ΔU=nC_{v,m}(T_{f}-T_{i})=-P_{external}(V_{f}-V_{i})$. for a non reversible adiabatic process. Im rather confused as to decide under which conditions should i use each of these three formulas. $\endgroup$ – wei jit Jan 26 '15 at 15:41

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