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Chapter 12: Problem 24

A system is taken from state \(A\) to state \(B\) along two different paths 1 and \(2 .\) The heat absorbed and work done by the system along these two paths are \(Q_{1}\) and \(Q_{2}\) and \(W_{1}\) and \(W_{2}\), respectively. (A) \(Q_{1}=Q_{2}\) (B) \(W_{1}=W_{2}\) (C) \(Q_{1}+W_{1}=Q_{2}+W_{2}\) (D) None of these

Short Answer

Expert verified
(C) \(Q_{1} + W_{1} = Q_{2} + W_{2}\) is the correct answer. Based on the First Law of Thermodynamics, the total energy change, which includes the heat absorbed by the system plus the work done by the system from going from state A to state B, must be the same regardless of the path and is equal to the internal energy change of the system, which is a constant.

Step by step solution

01

Understanding Concepts

The First Law of Thermodynamics is defined as \(\delta Q = dU + \delta W\) where \(\delta Q\) is the heat absorbed by the system, \(dU\) is the change in internal energy and \(\delta W\) is the work done by the system. In this case, the internal energy change would be the same, as the continuous state of the system, from state A to state B, doesn't depend on the path taken. The heat absorbed and the work done by the system may be different for different paths.
02

Applying the First Law of Thermodynamics for Path 1 and Path 2

For path 1, applying the first law of thermodynamics, we get: \(Q_{1} = \Delta U + W_{1}\) and for path 2, we get: \(Q_{2} = \Delta U + W_{2}\). Since the change in internal energy (\(\Delta U\)) must be the same for both paths (as it's a state function and doesn't depend on the path), we can set these two equations equal to each other.
03

Comparing the Equations

Setting the equations equal to each other, we get \(Q_{1} + W_{1} = Q_{2} + W_{2}\). So, the total energy changes (\(Q + W\)) for both paths must be equal.

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Most popular questions from this chapter

One mole of an ideal gas is enclosed in a cylinder fitted with a frictionless piston and occupies a volume of \(1.5\) litre at a pressure of \(1.2 \mathrm{~atm} .\) It is subjected to a process given by equation \(T=\alpha V^{2}, \gamma\) (adiabatic constant) for the gas \(=1.5\). Choose the wrong statement. Given \(R \alpha=80 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{lit}^{-2}(R=\) gas constant and \(\alpha\) is constant) (A) The \(P-V\) diagram of the process is a straight line. (B) The work done by the gas in increasing the volume of the gas to 9 litre is \(3150 \mathrm{~J}\). (C) The change in the internal energy of the gas is \(12600 \mathrm{~J}\). (D) The heat supplied to the gas in the process is 1575 J.

An ideal gas undergoes a quasi-static, reversible process in which its molar heat capacity \(C\) remains constant. If during this process the relation of pressure \(P\) and volume \(V\) is given by \(P V^{n}=\) constant, then \(n\) is given by (Here \(C_{P}\) and \(C_{V}\) are molar specific heat of constant pressure and constant volume, respectively) [2016] (A) \(n=\frac{C-C_{P}}{C-C_{V}}\) (B) \(n=\frac{C_{P}-C}{C-C_{V}}\) (C) \(n=\frac{C-C_{V}}{C-C_{P}}\) (D) \(n=\frac{C_{P}}{C_{V}}\)

The work of \(146 \mathrm{~kJ}\) is performed in order to compress 1 kilo mole of gas adiabatically and in this process, the temperature of the gas increases by \(7^{\circ} \mathrm{C}\). The gas is \(\left(R=8.3 \mathrm{~J} \mathrm{~mol}^{-1} K^{-1}\right)\) (A) Monoatomic (B) Diatomic(C) Triatomic (D) A mixture of monoatomic and diatomic

For an adiabatic expansion process, the quantity \(P V\) (A) decreases. (B) increases. (C) remains constant. (D) depends on adiabatic exponent of the gas.

\(n\) moles of a gas expands from volume \(V_{1}\) to \(V_{2}\) at constant temperature \(T\). The work done by the gas is (A) \(n R T\left(\frac{v_{2}}{v_{1}}\right)\) (B) \(n R T\left(\frac{v_{2}}{v_{1}}-1\right)\) (C) \(n R T \ln \left(\frac{v_{1}}{v_{2}}\right)\) (D) None of these
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