\(306 \mathrm{~J}\) of heat is required to raise the temperature of 2 moles of an ideal gas at constant…
\(306 \mathrm{~J}\) 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
\(306 \mathrm{~J}\)
\(153 \mathrm{~J}\)
\(140 \mathrm{~J}\)
\(80 \mathrm{~J}\)
Solution
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.
\(\begin{aligned}
Q_V & =n C_V \Delta T \\
& =2 \times 6.986 \times(35-25) \\
& =2 \times 6.986 \times 10 \\
& =139.72 \simeq 140 \mathrm{~J}
\end{aligned}\)