One kg of dry air at 15°C is isothermally compressed to one tenth of its initial volume. The work done on the system is .......... kJ. (Round off to the nearest integer.)[Assume that the gas constant for dry air is \(287 \times 10^5 \, {J K}^{-1} \, {kg}^{-1}\).]
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For isothermal processes, the work done is related to the change in volume and the temperature of the system. In an isothermal compression, the work is positive when the gas is compressed.
The work done during an isothermal compression process can be calculated using the following formula for an ideal gas:
\[
W = -n R T \ln\left(\frac{V_f}{V_i}\right)
\]
Where:
- \( W \) is the work done by or on the gas,
- \( n \) is the number of moles of the gas,
- \( R \) is the gas constant,
- \( T \) is the absolute temperature,
- \( V_f \) and \( V_i \) are the final and initial volumes, respectively.
Given that:
- The amount of dry air is 1 kg,
- The gas constant \( R = 287 \times 10^5 \, {J K}^{-1} \, {kg}^{-1} \),
- The temperature \( T = 15^\circ C = 15 + 273 = 288 \, {K} \),
- The volume is compressed to one-tenth of the initial volume, so \( \frac{V_f}{V_i} = \frac{1}{10} \).
Substituting these values into the formula:
\[
W = -(1 \, {kg}) \times (287 \times 10^5 \, {J K}^{-1} \, {kg}^{-1}) \times (288 \, {K}) \times \ln\left(\frac{1}{10}\right)
\]
\[
W = -(287 \times 10^5) \times 288 \times \ln(0.1)
\]
\[
W = -(287 \times 10^5) \times 288 \times (-2.3026)
\]
\[
W = 1.99 \times 10^8 \, {J} = 199 \, {kJ}
\]
