xpansion:

$$ W = -nRT \ln\left(\frac{V_2}{V_1}\right) $$

where:

• $ W $ is the work done by the gas (given as $-16.5\,\text{kJ}=-16500\,\text{J}$),

• $ n = 2.5\,\text{mol} $,

• $ R = 8.314\,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1} $,

• $ V_1 = 2\,\text{dm}^3 $ and $ V_2 = 20\,\text{dm}^3 $.

Follow these steps:

Write down the work formula:

$$ W = -nRT\ln\left(\frac{V_2}{V_1}\right) $$

Solve for the temperature $ T $:

$$ T = -\frac{W}{nR\ln\left(\frac{V_2}{V_1}\right)} $$

Substitute the known values:

$$ T = -\frac{-16500}{2.5 \times 8.314 \times \ln\left(\frac{20}{2}\right)} $$

Calculate the volume ratio and its natural logarithm:

$$ \frac{V_2}{V_1} = \frac{20}{2} = 10, \quad \text{so} \quad \ln(10) \approx 2.3026. $$

Now compute the temperature:

$$ T = \frac{16500}{2.5 \times 8.314 \times 2.3026}. $$

First, compute the product in the denominator:

$$ 2.5 \times 8.314 = 20.785, $$

and then

$$ 20.785 \times 2.3026 \approx 47.88. $$

Finally, calculate $ T $:

$$ T \approx \frac{16500}{47.88} \approx 344.8 \, \text{K}. $$

Rounding off to the nearest value gives approximately $ 345 \, \text{K} $.

Thus, the correct option is Option C: 345.

Thermodynamics

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