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MHT CET · Chemistry · Thermodynamics (C)

An ideal gas expands by \(1.5 \mathrm{~L}\) against a constant external pressure of \(2 \mathrm{~atm}\) at \(298 \mathrm{~K}\). Calculate the work done?

  1. A \(-75 \mathrm{~J}\)
  2. B \(-303.9 \mathrm{~J}\)
  3. C \(13.3 \mathrm{~J}\)
  4. D \(-30 \mathrm{~J}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(-303.9 \mathrm{~J}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{W} =-\mathrm{P}_{\text {ext }} \Delta \mathrm{V} \)
\( =-2 \mathrm{~atm} \times(1.5 \mathrm{~L}) \)
\( =-3 \mathrm{~atm} \mathrm{~L} \times 1.01325=-3.0398 \mathrm{dm}^3 \mathrm{bar}\)
Now, \(1 \mathrm{dm}^3\) bar \(=100 \mathrm{~J}\)
\(\text {Hence, } -3.0398 \mathrm{dm}^3 \text { bar } \times \frac{100 \mathrm{~J}}{1 \mathrm{dm}^3 \mathrm{bar}} =-303.98 \mathrm{~J} \)
\( \cong-303.9 \mathrm{~J}\)