Question 648: Thermodynamics & KTG

Question

The heat energy that must be supplied to 14 gram of nitrogen at room temperature to raise its temperature by $48^{\circ} \mathrm{C}$ at constant pressure is (Molecular weight of nitrogen $=28, R=$ gas constant, $\mathrm{C}_{\mathrm{p}}=\frac{7}{2} \mathrm{R}$ for diatomic gas)

Accepted Answer

Step 1: Recall formula for heat at constant pressure
$ Q = n C_p \Delta T $
where
$n = \frac{m}{M}$ = number of moles,
$C_p = 	frac{7}{2} R$ for diatomic gas,
$\Delta T = 48 \, K$.
Step 2: Calculate number of moles
Molecular weight of nitrogen ($N_2$) $M = 28$.
Mass given $m = 14 \, g$.
$ n = \frac{14}{28} = 0.5 \, 	ext{mol} $
Step 3: Substitute into equation
$ Q = n C_p \Delta T  = \left(0.5ight) \left(\frac{7}{2}Right) (48) $
Step 4: Simplify
$ Q = 0.5 	imes \frac{7}{2} 	imes 48 \, R $
$ Q = \frac{7}{4} 	imes 48 \, R $
$ Q = 7 	imes 12 \, R = 84 R $
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Final Answer: Option B (84 R)

Answer

76 R

Answer

84 R

Answer

90 R

Answer

96 R

Thermodynamics & KTG

The heat energy that must be supplied to 14 gram of nitrogen at room temperature to raise its temperature by 48^{\circ} \mathrm{C} at constant pressure is (Molecular weight of nitrogen =28, R= gas constant, \mathrm{C}_{\mathrm{p}}=\frac{7}{2} \mathrm{R} for diatomic gas)

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