Q. If the temperature of 1 mole of ideal gas is changed from $0℃$ to $100℃$ at constant pressure, then work done in the process is $(R = 8.3\ J/$Mole-Kelvin)
A) $8.3 \times 10^{- 3}\ J$
B) $8.3 \times 10^{- 2}\ J$
C) $8.3 \times 10^{2}\ J$
D) $8.3 \times 10^{3}\ J$
Explanation
(c) $W = P\mathrm{\Delta}V$
$nR(\mathrm{\Delta}T) = 1 \times 8.3 \times 100 = 8.3 \times 10^{2}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.