| ### Statement | | ### Statement |
| | | |
| $5.6.14.One mole of gas participating in the process, the graph of which is shown in | | $5.6.14.One mole of gas participating in the process, the graph of which is shown in |
| the figure, passes through states 1, 2, and 3 sequentially. The internal energy | | the figure, passes through states 1, 2, and 3 sequentially. The internal energy |
| of the gas is proportional to the temperature (U = cT). Find the amount of | | of the gas is proportional to the temperature (U = cT). Find the amount of |
| heat absorbed by the gas in this process. | | heat absorbed by the gas in this process. |
|  | |  |
| | | |
| ### Solution | | ### Solution |
| | | |
| Assuming we're working with ideal gases\ | | Assuming we're working with ideal gases\ |
| We have process 1-2 (isochoric)\ | | We have process 1-2 (isochoric)\ |
| $W_{12}=0$\ | | $W_{12}=0$\ |
| and for first law of thermodynamics\ | | and for first law of thermodynamics\ |
| $Q_{12}=\Delta U_{12}$\ | | $Q_{12}=\Delta U_{12}$\ |
| $Q_{12}=c(T_2 - T_1)$\ | | $Q_{12}=c(T_2 - T_1)$\ |
| and for process 2-3 (isobaric), we'll have\ | | and for process 2-3 (isobaric), we'll have\ |
| $Q_{23}=\Delta U_{23} + W_{23}$ , where $W_{23}=P_2(V_2 -V_1)$, $P_2=P_3$\ | | $Q_{23}=\Delta U_{23} + W_{23}$ , where $W_{23}=P_2(V_2 -V_1)$, $P_2=P_3$\ |
| @@ -19,7 +19,7 @@Solution |
| $Q_{23}=c(T_3 - T_2) + P_2(V_2 -V_1)$\ |
| $Q_{23}=c(T_3 - T_2) + P_2(V_2 -V_1)$\ |
| The amount of heat absorbed by the gas in this whole process is\ |
| The amount of heat absorbed by the gas in this whole process is\ |
| $Q_a=Q_{12 }+Q_{23}$\ |
| $Q_a=Q_{12 }+Q_{23}$\ |
| $Q_a=c(T_2 -T_1) + c(T_3 -T_2)+P_2(V_2 -V_1)$\ |
| $Q_a=c(T_2 -T_1) + c(T_3 -T_2)+P_2(V_2-V_1)$\ |
| $Q_a=c(T_3 -T_1)+P_2(V_2 -V_1)$\ |
| $Q_a=c(T_3 -T_1)+P_2(V_2 -V_1)$\ |
| by the ideal gas law, we have\ |
| by the ideal gas law, we have\ |
| $P_1V_1=RT_1$, and $P_2V_2=RT_3$, from this, we get\ |
| $P_1V_1=RT_1$, and $P_2V_2=RT_3$, from this, we get\ |
| $T_3 -T_1=\frac{P_2V_2 -P_1V_1}{R}$\ | | $T_3 -T_1=\frac{P_2V_2 -P_1V_1}{R}$\ |
| and substituting this into Q_a, we get\ | | and substituting this into Q_a, we get\ |
| $Q_a=\frac{c(P_2V_2 -P_1V_1)}{R} + P_2(V_2 -V_1)$ | | $Q_a=\frac{c(P_2V_2 -P_1V_1)}{R} + P_2(V_2 -V_1)$ |
| | | |
| #### Answer | | #### Answer |
| $Q_a=\frac{c(P_2V_2 -P_1V_1)}{R} +P_2(V_2 -V_1)$ | | $Q_a=\frac{c(P_2V_2 -P_1V_1)}{R} +P_2(V_2 -V_1)$ |
