When 300 J of heat given to an ideal gas with $C_p=\frac{7}{2} R$ its temperature raises from $20^{\

Question

When 300 J of heat given to an ideal gas with $C_p=\frac{7}{2} R$ its temperature raises from $20^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ keeping its volume constant. The amount of the gas is (approximately in millimoles) $\_\_\_\_$ g. $(\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} . \mathrm{K})$

Professor PC Thomas Classes · Answer

Because the process takes place at a constant volume, the appropriate specific heat capacity is $C_v$. Using Mayer's relation, $C_v = C_p - R = \frac{7}{2}R - R = \frac{5}{2}R$. The heat supplied is formulated as $Q = n C_v \Delta T$. Substituting the provided parameters: $300 = n (\frac{5}{2} 	imes 8.314) (50 - 20) = n 	imes 2.5 	imes 8.314 	imes 30 = 623.55 n$. Solving for the amount of substance yields $n = \frac{300}{623.55} \approx 0.481 	ext{ moles}$. This translates dimensionally to 481 millimoles. (Note: Due to a phrasing anomaly in the source question asking for "mass in grams" without specifying the molar mass, the numerical answer strictly evaluates the millimoles).