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Chapter 8: Problem 1119

The internal energy change in a system that has absorbed 2 Kcal of heat and done \(500 \mathrm{~J}\) of work is (A) \(7900 \mathrm{~J}\) (B) \(4400 \mathrm{~J}\) (C) \(6400 \mathrm{~J}\) (D) \(8900 \mathrm{~J}\)

Short Answer

Expert verified
The internal energy change in the system is 7868 J, which is closest to option (A) \(7900 \mathrm{~J}\).

Step by step solution

01

Convert heat absorbed from Kcal to Joules

To convert the heat absorbed from Kcal to Joules, we will use the conversion factor 1 Kcal = 4184 Joules. Heat absorbed = 2 Kcal x 4184 Joules/Kcal = 8368 Joules
02

Apply the first law of thermodynamics

According to the first law of thermodynamics, the change in internal energy of the system ΔU is given by: ΔU = Q - W where Q is the heat absorbed by the system and W is the work done by the system.
03

Calculate the change in internal energy

Now we can plug in the values for Q and W into the formula for ΔU: ΔU = 8368 Joules - 500 Joules ΔU = 7868 Joules
04

Compare the result with the given options

We have found that the change in internal energy ΔU is 7868 Joules. Since it is closest to 7900 Joules, the correct option is: (A) \(7900 \mathrm{~J}\)

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Most popular questions from this chapter

For free expansion of the gas which of the following is true ? (A) \(\mathrm{Q}=0, \mathrm{~W}>0\) and \(\mu \operatorname{Eint}=-\mathrm{W}\) (B) \(\mathrm{W}=0, \mathrm{Q}>0\) and \(\Delta\) Eint \(=\mathrm{Q}\) (C) \(\mathrm{W}>0, \mathrm{Q}<0\) and \(\Delta\) Eint \(=0\) (D) \(\mathrm{Q}=\mathrm{W}=0\) and \(\Delta \operatorname{Eint}=0\)

Air is filled in a motor tube at \(27^{\circ} \mathrm{C}\) and at a Pressure of 8 atmosphere. The tube suddenly bursts. Then what is the temperature of air. given \(\gamma\) of air \(=1.5\) (A) \(150 \mathrm{~K}\) (B) \(150^{\circ} \mathrm{C}\) (C) \(75 \mathrm{~K}\) (D) \(27.5^{\circ} \mathrm{C}\)

Work done by \(0.1\) mole of a gas at \(27^{\circ} \mathrm{C}\) to double its volume at constant Pressure is \(\quad\) Cal. $R=2\left\\{(\mathrm{Cal}) /\left(\mathrm{mol}^{\circ} \mathrm{K}\right)\right\\}$ (A) 600 (B) 546 (C) 60 (D) 54

\(\mu\) moles of gas expands from volume \(\mathrm{V}_{1}\) to \(\mathrm{V}_{2}\) at constant temperature \(\mathrm{T}\). The work done by the gas is (A) \(\mu \mathrm{RT}\left\\{\mathrm{V}_{2} / \mathrm{V}_{1}\right\\}\) (B) \(\mu \operatorname{RTln}\left\\{\mathrm{V}_{2} / \mathrm{V}_{1}\right\\}\) (C) $\mu \mathrm{RT}\left\\{\left(\mathrm{V}_{\mathrm{v}} / \mathrm{V}_{1}\right)-1\right\\}$ (D) $\mu \operatorname{RTln}\left\\{\left(\mathrm{V}_{2} / \mathrm{V}_{1}\right)\right.$

Efficiency of a Carnot engine is \(50 \%\), when temperature of outlet is $500 \mathrm{~K}\(. in order to increase efficiency up to \)60 \%$ keeping temperature of intake the same what is temperature of out let. (A) \(200 \mathrm{~K}\) (B) \(400 \mathrm{~K}\) (C) \(600 \mathrm{~K}\) (D) \(800 \mathrm{~K}\)
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