Q146 / 200 · WBJEE 2014 · Physics

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, $R=8.32 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ )

  • A
    $0.83 \times 10^{3} \mathrm{J}$
  • B
    $46 \times 10^{3} \mathrm{J}$
  • C
    $2.08 \times 10^{3} \mathrm{J}$
  • Correct answer
    D
    $1.25 \times 10^{3} \mathrm{J}$

Solution

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 $$ =1.25 \times 10^{3} \mathrm{J} $$

Hence, the correct answer is option: 'D'.

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