1. Tardigrade
  2. Question
  3. Physics
  4. If quantity of heat 1163.4 J supplied to one mole of nitrogen gas, at room temperature at constant pressure, then the rise in temperature is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If quantity of heat 1163.4 J supplied to one mole of nitrogen gas, at room temperature at constant pressure, then the rise in temperature is

KEAMKEAM 2008Kinetic Theory

Solution:

Heat given to the gas at room temperature and at constant temperature.
Q = $ n C_p \triangle \, T $
$\therefore 1163.4 = 1 \times \frac{ 7}{2} R \times \triangle T $
$ $ $ ( \because C_p = \frac{ 7}{2}$ R for diatomic gas)
or $ \triangle T = \frac{ 2 \times 1163.4 }{ 7 \times 8.31 } $
$ $ $ ( \because R = 8.31 \, J \, mol^{-1} \, K^{ - 1} )$
or $ \triangle T = 40 K $