Edit to “Statement”

Luisito edited Jun 16, 2026, 1:30 PM

View solution History

revision #18920 parent #18919 ← older newer →

en/5.6.14.md +1 −1

		@@ -1,6 +1,6 @@
			### 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
			of the gas is proportional to the temperature (U = cT). Find the amount of
			heat absorbed by the gas in this process.
			![Ð¡Ð½Ð¸Ð¼Ð¾Ðº ÑÐºÑÐ°Ð½Ð° 2025-06-18 160955.png\|459x311, 50%](../../img/5.6.14/Ð¡Ð½Ð¸Ð¼Ð¾Ðº ÑÐºÑÐ°Ð½Ð° 2025-06-18 160955.png)

			### Solution

			Assuming we're working with ideal gases\
			We have process 1-2 (isochoric)\
			$W_{12}=0$\
			and for first law of thermodynamics\
			$Q_{12}=\Delta U_{12}$\
			$Q_{12}=c(T_2 - T_1)$\
			and for process 2-3 (isobaric), we'll have\
			$Q_{23}=\Delta U_{23} + W_{23}$ , where $W_{23}=P_2(V_2 -V_1)$, $P_2=P_3$\
			$Q_{23}=c(T_3 - T_2) + P_2(V_2 -V_1)$\
			The amount of heat absorbed by the gas in this whole process is\
			$Q_a=Q_{12 }+Q_{23}$\
			$Q_a=c(T_2 -T_1) + c(T_3 -T_2)\+P_2(V_2-V_1)$\
			$Q_a=c(T_3 -T_1)+P_2(V_2 -V_1)$\
			by the ideal gas law, we have\
			$P_1V_1=RT_1$, and $P_2V_2=RT_3$, from this, we get\
			$T_3 -T_1=\frac{P_2V_2 -P_1V_1}{R}$\
			and substituting this into Q_a, we get\
			$Q_a=\frac{c(P_2V_2 -P_1V_1)}{R} + P_2(V_2 -V_1)$

			#### Answer
			$Q_a=\frac{c(P_2V_2 -P_1V_1)}{R} +P_2(V_2 -V_1)$
		unchanged lines 25

	@@ -1,6 +1,6 @@
	### 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.
	![Ð¡Ð½Ð¸Ð¼Ð¾Ðº ÑÐºÑÐ°Ð½Ð° 2025-06-18 160955.png\|459x311, 50%](../../img/5.6.14/Ð¡Ð½Ð¸Ð¼Ð¾Ðº ÑÐºÑÐ°Ð½Ð° 2025-06-18 160955.png)		![Ð¡Ð½Ð¸Ð¼Ð¾Ðº ÑÐºÑÐ°Ð½Ð° 2025-06-18 160955.png\|459x311, 50%](../../img/5.6.14/Ð¡Ð½Ð¸Ð¼Ð¾Ðº ÑÐºÑÐ°Ð½Ð° 2025-06-18 160955.png)

	### 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 $W_{23}=P_2(V_2 -V_1)$, $P_2=P_3$\
	$Q_{23}=c(T_3 - T_2) + P_2(V_2 -V_1)$\		$Q_{23}=c(T_3 - T_2) + P_2(V_2 -V_1)$\
	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
	$Q_a=\frac{c(P_2V_2 -P_1V_1)}{R} +P_2(V_2 -V_1)$		$Q_a=\frac{c(P_2V_2 -P_1V_1)}{R} +P_2(V_2 -V_1)$
	unchanged lines 25