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Thermodynamics — JEE physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceJEE physicsThermodynamics1 Mark
MCQ
A system changes from the state $\left(P_1, V_1\right)$ to $\left(P_2 V_2\right)$ as shown in the figure. What is the work done by the system
Image
  • A
    $7.5 \times 10^5$ joule
  • B
    $7.5 \times 10^5 \mathrm{erg}$
  • ✓
    $12 \times 10^5$ joule
  • D
    $6 \times 10^5$ joule
✓

Answer

Correct option: C.
$12 \times 10^5$ joule
Work done $=$ Area of $P V$ graph (here trapezium)
$=\frac{1}{2}\left(1 \times 10^5+5 \times 10^5\right) \times(5-1)=12 \times 10^5 J$

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