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One mole of an ideal monoatomic gas is heated at a constant pressure from $0^{\circ} \mathrm{C}$ to $100^\circ C$. Then the change in the internal enery of the gas is (Given, $R=8.32 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ )
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Verified Answer
The correct answer is:
$1.25 \times 10^{3} \mathrm{J}$
We know that $\Delta U=n C_{v} \Delta T$
$$
\begin{array}{l}
\text { where } \begin{aligned}
\Delta T &=\left(T_{2}-T_{1}\right) \\
T_{1} &=0^{\circ} \mathrm{C}=273 \mathrm{K} \\
T_{2} &=100^{\circ} \mathrm{C}=373 \mathrm{K} \\
n &=1 \text { (monoatomic gas) }
\end{aligned} \\
\begin{aligned}
R=8.32 \mathrm{J} \mathrm{mol}^{-} \mathrm{k}^{-1}=1 \times \frac{3}{2} R \times(373-273) \\
=1 \times \frac{3}{2} \times 8.32 \times 100=3 \times 8.32 \times 50
\end{aligned}
\end{array}
$$
$\Rightarrow 1248 \mathrm{kJ}$
of
$$
=1.25 \times 10^{3} \mathrm{J}
$$
$$
\begin{array}{l}
\text { where } \begin{aligned}
\Delta T &=\left(T_{2}-T_{1}\right) \\
T_{1} &=0^{\circ} \mathrm{C}=273 \mathrm{K} \\
T_{2} &=100^{\circ} \mathrm{C}=373 \mathrm{K} \\
n &=1 \text { (monoatomic gas) }
\end{aligned} \\
\begin{aligned}
R=8.32 \mathrm{J} \mathrm{mol}^{-} \mathrm{k}^{-1}=1 \times \frac{3}{2} R \times(373-273) \\
=1 \times \frac{3}{2} \times 8.32 \times 100=3 \times 8.32 \times 50
\end{aligned}
\end{array}
$$
$\Rightarrow 1248 \mathrm{kJ}$
of
$$
=1.25 \times 10^{3} \mathrm{J}
$$
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