Q. If a quantity of heat 1163.4 joule is supplied to one mole of nitrogen gas, at room temperature at constant pressure, then the rise in temperature is (Given $R = 8.31J\ mole^{- 1}K^{- 1}$)
A) $54K$
B) $28K$
C) $65K$
D) $40K$
Explanation
(d) $\mathrm{\Delta}Q = \mu C_{P}dT$
$\Rightarrow 1163.4 = 1.\frac{7}{2}R.dT$
$\Rightarrow dT = \frac{1163.4 \times 2}{7 \times 8.31} = 40K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.