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A $\mathrm{200.0\ m^3}$ balloon at $253.0\ \mathrm K$ ascends to a higher altitude. If the initial pressure inside the balloon is $325$ millibar, and it ascends to a latitude with a pressure of $7.45$ millibar, by what percent does the absolute temperature change:

  1. Assuming the process is reversible and adiabatic
  2. Assuming the process is irreversible and adiabatic

All of the formulae I have on this subject include moles. The question does not give a value of moles. Where do I start?

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  • $\begingroup$ Try to use the formulas you have to calculate the moles. $\endgroup$ – Curt F. Sep 30 '15 at 3:33
  • $\begingroup$ I assume that you have to assume that the gas behaves ideally. $\endgroup$ – orthocresol Sep 30 '15 at 5:55
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The change in internal energy $U$ is $$\Delta U=Q+W$$ where $Q$ is amount of heat transferred to the system and $W$ is work done on the system.
Since the process is adiabatic, no heat is transferred into or out of the system, i.e. $Q=0$ and thus $$\Delta U=W$$

The reversible expansion is performed continuously at equilibrium by means of infinitesimal changes in pressure $p$.
The corresponding infinitesimal pressure-volume work $\delta W$ is $$\delta W=-p\,\mathrm dV$$

For an ideal gas, the internal energy $U$ only depends on amount of substance $n$ and temperature $T$. Since $n$ is constant in a closed system, the change in internal energy $U$ is $$\mathrm dU=C_V\,\mathrm dT$$ where $C_V$ is heat capacity at constant volume.

Equating $\mathrm dU$ and $\delta W$ yields $$C_V\,\mathrm dT=-p\,\mathrm dV$$ and since $pV=nRT$ for an ideal gas $$\begin{align} C_V\,\mathrm dT&=-\frac{nRT}V\,\mathrm dV\\ \frac{C_V}T\,\mathrm dT&=-\frac{nR}V\,\mathrm dV \end{align}$$

Integration from the initial state ($T_1,V_1$) to the final state ($T_2,V_2$) yields $$\begin{align} \int\limits_{T_1}^{T_2}\frac{C_V}T\,\mathrm dT&=\int\limits_{V_1}^{V_2}-\frac{nR}V\,\mathrm dV\\ C_V\int\limits_{T_1}^{T_2}\frac1T\,\mathrm dT&=-nR\int\limits_{V_1}^{V_2}\frac1V\,\mathrm dV\\ C_V\ln\left(\frac{T_2}{T_1}\right)&=-nR\ln\left(\frac{V_2}{V_1}\right)\\ \ln\left(\frac{T_2}{T_1}\right)&=-\frac{nR}{C_V}\ln\left(\frac{V_2}{V_1}\right)\\ \frac{T_2}{T_1}&=\left(\frac{V_2}{V_1}\right)^{-nR/C_V}\\ &=\left(\frac{V_1}{V_2}\right)^{R/C_{\mathrm m,V}} \end{align}$$ where $C_{\mathrm m,V}$ is molar heat capacity at constant volume.

Since $C_{\mathrm m,p}-C_{\mathrm m,V}=R$, the exponent can be rewritten $$\begin{align} \frac{T_2}{T_1}&=\left(\frac{V_1}{V_2}\right)^{R/C_{\mathrm m,V}}\\ &=\left(\frac{V_1}{V_2}\right)^{C_{\mathrm m,p}/C_{\mathrm m,V}-1}\\ &=\left(\frac{V_1}{V_2}\right)^{\gamma-1}\\ \end{align}$$ where $\gamma$ is the ratio of the heat capacities $\gamma=C_p/C_V=C_{\mathrm m,p}/C_{\mathrm m,V}=c_p/c_V$. For a monoatomic ideal gas, $C_{\mathrm m,p}=\tfrac52R$ and $C_{\mathrm m,V}=\tfrac32R$, and thus $\gamma=\tfrac53$.

Note that, in the given question, the volumes $V_1$ and $V_2$ are unknown. However, the values for the pressures $p_1$ and $p_2$ are given. Since $pV=nRT$ for an ideal gas, $$\begin{align} \frac{T_2}{T_1}&=\left(\frac{V_1}{V_2}\right)^{\gamma-1}\\ T_2\cdot V_2^{\gamma-1}&=T_1\cdot V_1^{\gamma-1}\\ \frac{p_2V_2}{nR}\cdot V_2^{\gamma-1}&=\frac{p_1V_1}{nR}\cdot V_1^{\gamma-1}\\ p_2\cdot V_2\cdot V_2^{\gamma-1}&=p_1\cdot V_1\cdot V_1^{\gamma-1}\\ p_2\cdot V_2^\gamma&=p_1\cdot V_1^\gamma\\ \frac{p_2}{p_1}&=\left(\frac{V_1}{V_2}\right)^\gamma \end{align}$$

In contrast to the reversible expansion, an irreversible expansion is not performed continuously at equilibrium by means of infinitesimal changes in pressure. In the limiting case, the value of the pressure $p$ abruptly changes from $p_1$ and $p_2$. The following expansion proceeds in one step against a constant pressure $p_2$. Hence, $$\delta W=-p_2\,\mathrm dV$$ and thus $$\begin{align} C_V\,\mathrm dT&=-p_2\,\mathrm dV\\ \int\limits_{T_1}^{T_2}C_V\,\mathrm dT&=\int\limits_{V_1}^{V_2}-p_2\,\mathrm dV\\ C_V\left(T_2-T_1\right)&=-p_2\left(V_2-V_1\right) \end{align}$$ Since $pV=nRT$ for an ideal gas, $$\begin{align} C_V\left(T_2-T_1\right)&=-p_2\left(V_2-V_1\right)\\ &=-p_2\left(\frac{nRT_2}{p_2}-\frac{nRT_1}{p_1}\right)\\ &=-nRT_2+nRT_1\frac{p_2}{p_1}\\ C_VT_2+nRT_2&=C_VT_1+nRT_1\frac{p_2}{p_1}\\ T_2\left(C_V+nR\right)&=T_1\left(C_V+nR\frac{p_2}{p_1}\right)\\ T_2\left(C_{\mathrm m,V}+R\right)&=T_1\left(C_{\mathrm m,V}+R\frac{p_2}{p_1}\right)\\ \end{align}$$

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  • $\begingroup$ Why is U=nCvdT for ideal gas...... It should be valid only for isochoric yet this is used every process $\endgroup$ – Abhinav Aug 14 '18 at 10:17
  • $\begingroup$ @Abhinav $du =n*C_v*dT$ is valid for an ideal gas, irrespective of what process we are talking about. It is a fact for any ideal gas. $\endgroup$ – McSuperbX1 Aug 8 at 12:04
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where Cm,VCm,V is molar heat capacity at constant volume.

Since Cm,p−Cm,V=RCm,p−Cm,V=R, the exponent can be rewritten T2T1=(V1V2)R/Cm,V=(V1V2)Cm,p/Cm,V−1=(V1V2)γ−1 T2T1=(V1V2)R/Cm,V=(V1V2)Cm,p/Cm,V−1=(V1V2)γ−1 where γγ is the ratio of the heat capacities γ=Cp/CV=Cm,p/Cm,V=cp/cVγ=Cp/CV=Cm,p/Cm,V=cp/cV. For a monoatomic ideal gas, Cm,p=52RCm,p=52R and Cm,V=32RCm,V=32R, and thus γ=53γ=53.

Note that, in the given question, the volumes V1V1 and V2V2 are unknown. However, the values for the pressures p1p1 and p2p2 are given. Since pV=nRTpV=nRT for an ideal gas, T2T1T2⋅Vγ−12p2V2nR⋅Vγ−12p2⋅V2⋅Vγ−12p2⋅Vγ2p2p1=(V1V2)γ−1=T1⋅Vγ−11=p1V1nR⋅Vγ−11=p1⋅V1⋅Vγ−11=p1⋅Vγ1=(V1V2)γ T2T1=(V1V2)γ−1T2⋅V2γ−1=T1⋅V1γ−1p2V2nR⋅V2γ−1=p1V1nR⋅V1γ−1p2⋅V2⋅V2γ−1=p1⋅V1⋅V1γ−1p2⋅V2γ=p1⋅V1γp2p1=(V1V2)γ In contrast to the reversible expansion, an irreversible expansion is not performed continuously at equilibrium by means of infinitesimal changes in pressure. In the limiting case, the value of the pressure pp abruptly changes from p1p1 and p2p2. The following expansion proceeds in one step against a constant pressure p2p2. Hence, δW=−p2dV

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    $\begingroup$ Welcome to Chemistry.SE! Please go through this page to learn how to format your posts containing math expressions. It may look intimidating at first, but is really straightforward, and enhances the readablilty of your posts greatly! $\endgroup$ – William R. Ebenezer May 18 at 5:29

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