$306 \mathrm{~J}$ of heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from $25^{\circ} \mathrm{C}$ to $35^{\circ} \mathrm{C}$. The amount of heat required to raise the temperature of the same gas through the same range at constant volume is

Question

$306 \mathrm{~J}$ of heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from $25^{\circ} \mathrm{C}$ to $35^{\circ} \mathrm{C}$. The amount of heat required to raise the temperature of the same gas through the same range at constant volume is, $306 \mathrm{~J}$, $153 \mathrm{~J}$, $140 \mathrm{~J}$, $80 \mathrm{~J}$

MARKS App · Accepted Answer

Given, at constant pressure heat $Q_p=306 \mathrm{~J}$ Number of mole, $n=2$ $\begin{aligned} \Delta T & =T_2-T_1 \ & =35-25=10^{\circ} \mathrm{C} \end{aligned}$ We know that, $Q_p=n C_p \Delta T$ $\begin{array}{ll} \Rightarrow & 306=2 C_p \times 10 \ \Rightarrow & C_p=\frac{306}{2 \times 10} \ \Rightarrow & C_p=15.3 \mathrm{~J} / \mathrm{mol} \mathrm{K} \end{array}$ According to Mayer's formula, $\begin{aligned} & C_p-C_V =R \ \Rightarrow & C_V =C_p-R=15.3-8.314 \ \Rightarrow & C_V =6.986 \mathrm{~J} / \mathrm{mol} \mathrm{K} \end{aligned}$ The amount of heat required to raise the temperature of gas at constant volume through same range of temperature. $\begin{aligned} Q_V & =n C_V \Delta T \ & =2 \times 6.986 \times(35-25) \ & =2 \times 6.986 \times 10 \ & =139.72 \simeq 140 \mathrm{~J} \end{aligned}$

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