Question

if quantity of heat 1163.4 J supplied to one mole of nitrogen gas , at room temperture at constant pressure , then the rise in temperature is
A. 54 K
B. 28 K
C. 65 K
D. 8 K

Answer 1

Correct Answer - D
Heat given to the gas at room temperture and at constant temperature , `Q=nCpDeltaT`
`therefore1163.4=1xx(7)/(2)RxxDeltaT" " (becauseC_(p)=(7)/(2)R " for diatomic gas")`
`"or"" " DeltaT=(2xx1163.4)/(7xx8.31)" " (because R=8.31J"mol"^(-1)K^(-1))`
`"or " DeltaT=40K`