Question

How much energy in joules must be supplied to \[14{\rm{ g}}\] of nitrogen at room temperature to raise its temperature by \[40^\circ {\rm{ C}}\] at constant pressure? (Mol. wt of \[{N_2} = 28{\rm{ g}}\], \[R = \]constant)

Answer 1

Hint:As it is said that the temperature of nitrogen is to be raised by \[40^\circ {\rm{ C}}\] so we can say that we have to supply sensible heat which is expressed in the form of mass of nitrogen, specific heat of nitrogen and the amount of temperature that has to be raised.

Complete step by step answer:
Given:
Mass of the nitrogen is \[m = 14{\rm{ g}} = 0.014{\rm{ kg}}\].
The rise in temperature is \[\Delta T = 40^\circ {\rm{ C}} = 40{\rm{ K}}\].
Molecular weight of nitrogen per mole is \[M = 28{\rm{ g}} = 0.028{\rm{ kg}}\].
We have to evaluate the amount of heat required to raise the temperature of nitrogen by \[40^\circ {\rm{ C}}\].
The value of the gas constant of nitrogen can be expressed as the ratio of universal gas constant to the molecular weight of nitrogen.
\[R = \dfrac{{\bar R}}{M}\]……(1)
Universal gas constant is also known as Boltzmann constant and its value is given as:
\[\bar R = 8.314{\rm{ J mo}}{{\rm{l}}^{ - 1}}\]
Substitute \[8.314{\rm{ J mo}}{{\rm{l}}^{ - 1}}\] for \[\bar R\]and \[0.028{\rm{ kg}}\]for M in equation (1) to find the gas constant of nitrogen.
\[
R = \dfrac{{8.314{\rm{ J }}{{\rm{K}}^{ - 1}}{\rm{mo}}{{\rm{l}}^{ - 1}}}}{{0.028{\rm{ kgmo}}{{\rm{l}}^{ - 1}}}}\\
\Rightarrow R = 296.928 \times {10^{ - 3}}{\rm{ }}{{{\rm{kJ}}} {\left/
 {\vphantom {{{\rm{kJ}}} {{\rm{kgK}}}}} \right.
 } {{\rm{kgK}}}}
\]
We know that the specific heat at constant pressure for nitrogen is given as:
\[{c_p} = \dfrac{7}{2}R\]
Substitute \[296.928{\rm{ }}{{{\rm{kJ}}} {\left/
 {\vphantom {{{\rm{kJ}}} {{\rm{kgK}}}}} \right.
 } {{\rm{kgK}}}}\] for R in the above expression.
\[
{c_p} = \dfrac{7}{2}\left( {296.928{\rm{ }}{{{\rm{kJ}}} {\left/
 {\vphantom {{{\rm{kJ}}} {{\rm{kgK}}}}} \right.
 } {{\rm{kgK}}}}} \right)\\
\Rightarrow{c_p} = 1.039{\rm{ }}{{{\rm{kJ}}} {\left/
 {\vphantom {{{\rm{kJ}}} {{\rm{kgK}}}}} \right.
 } {{\rm{kgK}}}}
\]
Writing the expression for heat to be supplied to raise the temperature of nitrogen by \[\Delta T\].
\[Q = m{c_p}\Delta T\]
Substitute \[0.014{\rm{ kg}}\] for m, \[1.039{\rm{ }}{{{\rm{kJ}}} {\left/
{\vphantom {{{\rm{kJ}}} {{\rm{kgK}}}}} \right.
} {{\rm{kgK}}}}\] for \[{c_p}\] and \[40^\circ {\rm{ C}}\] for \[\Delta T\] in the above expression.
\[
Q = \left( {0.014{\rm{ kg}}} \right)\left( {1.039{\rm{ }}{{{\rm{kJ}}} {\left/
{\vphantom {{{\rm{kJ}}} {{\rm{kgK}}}}} \right.
} {{\rm{kgK}}}}} \right)\left( {40{\rm{ K}}} \right)\\
\therefore Q = 581.84{\rm{ J}}
\]

Therefore, the amount of heat required to raise the temperature of nitrogen by \[40^\circ {\rm{ C}}\] is equal to \[581.84{\rm{ J}}\].

Note: In the question molecular mass of nitrogen is given in gram but we have to understand that molecular mass of nitrogen is always per unit mole of nitrogen. Take extra care while performing unit conversion from CGS system to MKS system or vice versa.