When 1 , mol of nitrogen gas is heated at con | vedclass

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(D) For an isobaric process,the heat supplied is given by the formula: $dQ = \mu C_P dT$.Here,$\mu = 1 \, mol$,$dQ = 1163.4 \, J$,and $R = 8.31 \, J \

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(D) For an isobaric process,the heat supplied is given by the formula: $dQ = \mu C_P dT$.Here,$\mu = 1 \, mol$,$dQ = 1163.4 \, J$,and $R = 8.31 \, J \, mol^{-1} \, K^{-1}$.For a diatomic gas like nitrogen,the molar heat capacity at constant pressure is $C_P = \frac{7}{2} R$.Substituting the values into the equation: $dT = \frac{dQ}{\mu C_P}$.$dT = \frac{1163.4}{1 	imes (\frac{7}{2} 	imes 8.31)}$.$dT = \frac{1163.4 	imes 2}{7 	imes 8.31} = \frac{2326.8}{58.17} \approx 40 \, K$.Therefore,the increase in temperature is $40 \, K$.

When $1 \, mol$ of nitrogen gas is heated at constant pressure at room temperature,it is supplied with $1163.4 \, J$ of energy. The increase in temperature is ....... $K$ $(R = 8.31 \, J \, mol^{-1} \, K^{-1})$.

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