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NEET 2013
Exam
NEET
Year
2013
Subject
Physics
Chapter
Thermodynamics

NEET 2013 · Physics · Thermodynamics

A system is taken from state a to state c by two paths a d c and a...

A system is taken from state a to state cc by two paths adca d c and abca b c as shown in the figure. The internal energy a is U2=10U_2=10 J. Along the path adc the amount of heat heat absorbed δQ1=50 J\delta Q_1=50 \mathrm{~J} and the work obtained δW1=20 J\delta W_1=20 \mathrm{~J} whereas along the path abca b c the heat absorbed δQ2=36 J\delta Q_2=36 \mathrm{~J}. The amount of work along the path abca b c is: A system is taken from state a to state c by two paths a d c and a b c as shown in the figure. The internal energy a is

Options

  1. A10 J10 \mathrm{~J}
  2. B12 J12 \mathrm{~J}
  3. C38 J38 \mathrm{~J}
  4. D6 J6 \mathrm{~J}

Correct answer

D. 6 J6 \mathrm{~J}

Step-by-step solution

According to 1st 11^{\text{st }} 1 aw of thermodynamics δQ=δV+δW\delta Q=\delta V+\delta W Along the path adc δU1=δQ1−δW1=50−20=30JthenpathabcSincechangeinInternalenergyinpathindependence\delta U_1=\delta Q_1-\delta W_1=50-20=30 \mathrm{ J} then path abc Since change in Internal energy in path in dependenceSo, δU1=δU230J=36J−δW2⇒δW2=36J−30J=6J\begin{aligned}\text{So, } \delta U_1 & =\delta U_2 \\ 30 \mathrm{ J} & =36 \mathrm{ J}-\delta W_2 \\ \Rightarrow \delta W_2 & =36 \mathrm{ J}-30 \mathrm{ J} \\ & =6 \mathrm{ J} \end{aligned}$$
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