/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 83 A diatomic ideal gas is used in ... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 83

A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion, part of the cycle and the volume of the gas increases from \(\mathrm{V}\) to \(32 \mathrm{~V}\), the efficiency of the engine is [2010] (A) \(0.25\) (B) \(0.5\) (C) \(0.75\) (D) \(0.99\)

Short Answer

Expert verified
The efficiency of the Carnot engine during the adiabatic expansion when the volume of the gas increases from V to 32V is approximately \(0.5\) or \(50\%\). This is calculated using the formulas for adiabatic expansion and the properties of a diatomic ideal gas to find the temperatures of the hot and cold reservoirs, and then using the efficiency formula for Carnot engine. Therefore, the correct answer is (B) 0.5.

Step by step solution

01

For a diatomic ideal gas, the adiabatic expansion process can be described using the formula: \(P_1V_1^{\gamma} = P_2V_2^{\gamma}\) where \(P_1\) and \(P_2\) are the initial and final pressures, \(V_1\) and \(V_2\) are the initial and final volumes, and \(\gamma\) is the heat capacity ratio (\(\gamma = \frac{C_P}{C_V} = \frac{7}{5} \) for a diatomic ideal gas). We are given that \(V_2 = 32V_1\), so the formula becomes: \(P_1V_1^{\frac{7}{5}} = P_2(32V_1)^{\frac{7}{5}}\) #Step 2: Relate the temperatures and pressures#

We can relate the temperatures and pressures of the gas using the ideal gas law: \(P_1V_1 = nR T_1\) \(P_2V_2 = nR T_2\) where \(n\) is the number of moles of gas, \(R\) is the gas constant, and \(T_1\) and \(T_2\) are the temperatures before and after expansion, respectively. Dividing the second equation by the first equation, we get: \(\frac{P_2}{P_1} = \frac{T_2}{T_1}\frac{V_2}{V_1}\) Substitute \(V_2 = 32V_1\): \(\frac{P_2}{P_1} = 32\frac{T_2}{T_1}\) #Step 3: Solve for the temperature ratio#
02

Now, we substitute the expression for \(\frac{P_2}{P_1}\) from Step 1 into Step 2, and solve for the temperature ratio: \(32\frac{T_2}{T_1} = \left(\frac{V_1}{32V_1}\right)^{\frac{7}{5}}\) \(\frac{T_2}{T_1} = \frac{1}{32^{\frac{2}{5}}}\) #Step 4: Calculate the efficiency#

Using the efficiency formula for Carnot engine, we substitute the temperature ratio and find the efficiency: Efficiency = \(1 - \frac{T_2}{T_1} = 1 - \frac{1}{32^{\frac{2}{5}}}\) Efficiency 鈮� 0.5 So, the efficiency of the engine is about 0.5 or 50%. Therefore, the correct answer is (B) 0.5.

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Most popular questions from this chapter

One mole of an ideal monatomic gas is taken from temperature \(T_{0}\) to \(2 T_{0}\) by the process \(P T^{-4}=C\) Considering the following statements, choose the correct alternative. I. Molar heat capacity of the gas is \(-\frac{3 R}{2}\) II. Molar heat capacity of the gas is \(\frac{3 R}{2}\) III. Work done is \(-3 R T_{0}\) IV. Work done is \(3 R T_{0}\) (A) Statements I and IV are correct. (B) Statements I and III are correct. (C) Statements II and IV are correct. (D) Statements II and III are correct.

A gas expands under constant pressure \(P\) from volume \(V_{1}\) to \(V_{2}\). The work done by the gas is (A) \(P\left(V_{2}-V_{1}\right)\) (B) \(P\left(V_{1}-V_{2}\right)\) (C) \(P\left(V_{1}^{\gamma}-V_{2}^{\gamma}\right)\) (D) \(\frac{P V_{1} V_{2}}{V_{2}-V_{1}}\)

Three perfect gases at absolute temperatures, \(T_{1}, T_{2}\), and \(T_{3}\), are mixed. The masses of molecules are \(m_{1}, m_{2}\), and \(m_{3}\) and the number of molecules are \(n_{1}, n_{2}\), and \(n_{3}\), respectively. Assuming no loss of energy, the final temperature of the mixture is [2011] (A) \(\frac{\left(T_{1}+T_{2}+T_{3}\right)}{3}\) (B) \(\frac{n_{1} T_{1}+n_{2} T_{2}+n_{3} T_{3}}{n_{1}+n_{2}+n_{3}}\) (C) \(\frac{n_{1} T_{1}^{2}+n_{2} T_{2}^{2}+n_{3} T_{3}^{3}}{n_{1} T_{1}+n_{2} T_{2}+n_{3} T_{3}}\) (D) \(\frac{n_{1}^{2} T_{1}^{2}+n_{2}^{2} T_{2}^{2}+n_{3}^{3} T_{3}^{3}}{n_{1} T_{1}+n_{2} T_{2}+n_{3} T_{3}}\)

If amount of heat given to a system is \(50 \mathrm{~J}\) and work done on the system is \(15 \mathrm{~J}\), then change in internal energy of the system is (A) \(35 \mathrm{~J}\) (B) \(50 \mathrm{~J}\) (C) \(65 \mathrm{~J}\) (D) \(15 \mathrm{~J}\)

Three copper blocks of masses \(M_{1}, M_{2}\), and \(M_{3} \mathrm{~kg}\), respectively, are brought in to thermal contact till they reach equilibrium. Before contact, they were at \(T_{1}, T_{2}, T_{3}\left(T_{1}>T_{2}>T_{3}\right)\). Assuming there is no heat loss to the surroundings, the equilibrium temperature \(T\) is \((s\) is specific heat of copper) (A) \(T=\frac{T_{1}+T_{2}+T_{3}}{3}\) (B) \(T=\frac{M_{1} T_{1}+M_{2} T_{2}+M_{3} T_{3}}{M_{1}+M_{2}+M_{3}}\) (C) \(T=\frac{M_{1} T_{1}+M_{2} T_{2}+M_{3} T_{3}}{3\left(M_{1}+M_{2}+M_{3}\right)}\) (D) \(T=\frac{M_{1} T_{1} s+M_{2} T_{2} s+M_{3} T_{3} s}{M_{1}+M_{2}+M_{3}}\)
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