SATHEE: Heat and Thermodynamics 5 Question 2

Heat and Thermodynamics 5 Question 2

2. A sample of an ideal gas is taken through the cyclic process $a b c a$ as shown in the figure. The change in the internal energy of the gas along the path $c a$ is $- 180 J$ . The gas absorbs $250 J$ of heat along the path $a b$ and $60 J$ along the path $b c$ . The work done by the gas along the path $a b c$ is

(2019 Main, 12 April I)

(a) $120 J$

(b) $130 J$

(d) $140 J$

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Solution:

Key Idea In $p - V$ curve, work done $d W$ , change in internal energy $Δ U$ and heat absorbed $Δ Q$ are connected with first law of thermodynamics, i.e.

$Δ Q = Δ U + d W$

and total change in internal energy in complete cycle is always zero. Using this equation in different part of the curve, we can solve the

given problem.

In Process $a \to b$

$\begin{aligned} Given, & Δ Q_{a b} & = 250 J \\ ∴ & 250 J & = Δ U_{a b} + d W_{a b} \end{aligned}$

In Process $b \to c$

$Given, Δ Q_{b c} = 60 J$

Also, $V$ is constant, so $d V = 0$

$\begin{aligned} \Rightarrow & d W_{b c} & = p (d V)_{b c} = 0 \\ ∴ & 60 J & = Δ U_{b c} + 0 \\ \Rightarrow & Δ U_{b c} & = 60 J \end{aligned}$

In Process $c \to a$

Given, $Δ U_{c a} = - 180 J$

Now, for complete cycle,

$Δ U_{a b c a} = Δ U_{a b} + Δ U_{b c} + Δ U_{c a} = 0$

From Eqs. (iii), (iv) and (v), we get

$\begin{aligned} Δ U_{a b} & = - Δ U_{b c} - Δ U_{c a} \\ Δ U_{a b} & = - 60 + 180 = 120 J \end{aligned}$

From Eq. (ii), we get

$\begin{aligned} 250 J & = 120 J + d W_{a b} \\ \Rightarrow & d W_{a b} & = 130 J \end{aligned}$

From Eqs. (i) and (vii), we get

Work done by the gas along the path $a b c$ ,

$\begin{aligned} d W_{a b c} & = d W_{a b} + d W_{b c} \\ = 130 J + 0 K \\ \Rightarrow d W_{a b c} & = 130 J \end{aligned}$

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Heat and Thermodynamics 5 Question 2

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

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अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Heat and Thermodynamics 5 Question 2

2. A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is −180J. The gas absorbs 250J of heat along the path ab and 60J along the path bc. The work done by the gas along the path abc is

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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