Thermodynamic processes

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Question

n moles of an ideal gas undergo a thermodynamic process in which pressure of gas varies linearly with the volume of gas as shown in figure. If $\frac{P_{0} V_{0}}{nR} = 200 K$ , find the maximum temperature of the gas during the process (in K).

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Solution

Gas pressure is given as
$\begin{array}{l} S l o p e = \frac{2 P_{0} - P_{0}}{V_{0} - 2 V_{0}} = - \frac{P_{0}}{V_{0}} \\ (P - 2 P_{0}) = - \frac{P_{0}}{V_{0}} (V - V_{0}) \\ \Rightarrow P = 3 P_{0} - \frac{P_{0}}{V_{0}} V \\ T = \frac{PV}{nR} = \frac{3 P_{0}}{nR} V - \frac{P_{0}}{{nRV}_{0}} V^{2} \end{array}$
T is maximum when $\frac{dT}{dV} = 0$
$\begin{array}{l} \frac{dT}{dV} = \frac{3 P_{0}}{nR} - \frac{2 P_{0}}{{nRV}_{0}} V = 0 or, V = \frac{3 V_{0}}{2} \\ \Rightarrow T_{max} = \frac{3 P_{0}}{nR} (\frac{3 V_{0}}{2}) - \frac{P_{0}}{{nRV}_{0}} {(\frac{3}{2} V_{0})}^{2} \\ = \frac{9}{2} \frac{P_{0} V_{0}}{nR} - \frac{9}{4} \frac{P_{0} V_{0}}{nR} = \frac{9 P_{0} V_{0}}{4 nR} \\ \Rightarrow T_{max} = \frac{9}{4} \times 200 = 450 K \end{array}$

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Thermodynamic processes

Remember concepts with our Masterclasses.

n moles of an ideal gas undergo a thermodynamic process in which pressure of gas varies linearly with the volume of gas as shown in figure. If P0V0nR=200K, find the maximum temperature of the gas during the process (in K).

Gas pressure is given as Slope=2P0-P0V0-2V0=-P0V0 P−2P0=−P0V0V−V0⇒P=3P0−P0V0V T=PVnR=3P0nRV−P0nRV0V2T is maximum when dTdV=0 dTdV=3P0nR−2P0nRV0V=0 or, V=3V02⇒ Tmax=3P0nR3V02−P0nRV032V02 =92P0V0nR−94P0V0nR=9P0V04nR⇒ Tmax=94×200=450K

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n moles of an ideal gas undergo a thermodynamic process in which pressure of gas varies linearly with the volume of gas as shown in figure. If $\frac{P_{0} V_{0}}{nR} = 200 K$ , find the maximum temperature of the gas during the process (in K).