An ideal gas expands from 100 cm^3 to 200 cm^3 at a constant...

Work done by the gas is given by:

$\Delta W = P \Delta V = 2.0 \times 10^5 \times (200 - 100) \times 10^{-6} = 20 \text{ J}$

From the first law of thermodynamics:

$\Delta Q = \Delta U + \Delta W$

$50 = \Delta U + 20 \Rightarrow \Delta U = 30 \text{ J}$

The change in internal energy is:

$\Delta U = n C_V \Delta T$

For an isobaric process, the ideal gas equation gives $P \Delta V = n R \Delta T$. Thus:

$n \Delta T = \dfrac{P \Delta V}{R} = \dfrac{20}{R}$

Substituting this into the equation for $\Delta U$:

$30 = C_V$\left( \dfrac{20}{R} \right)<br><br>C_V = $\dfrac{30 R}{20} =$1.5 R$<br><br>Using the universal gas constant$R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$:<br><br>$C_V = 1.5 \times 8.314 \approx 12.5 \text{ J K}^{-1} \text{ mol}^{-1}