For an isothermal irreversible expansion, Work done (W)=−pext(V2−V1) where, V1= initial volume V2= final volume Given, pext=105Nm−2, V1=10−3 m3, V2=10−2 m3 On substituting the given values in Eq, (i), We get, W=−105Nm−2(10−2 m3−10−3 m3)=−105Nm−2×10−3(10−1)m3=−900Nm=−900 J
An ideal gas expands isothermally from 10^-3 ~m^3 to 10^-2 ~m^3 at 300 ~K against a constant pressure of 10^5 Nm^-2. The work done on the gas is
NEET UG 2019 — Chemistry Physical Chemistry
2019mcqmedium
An ideal gas expands isothermally from 10−3 m3 to 10−2 m3 at 300 K against a constant pressure of 105Nm−2. The work done on the gas is
Official previous-year question
Held on 30 Apr 2019 · Verified 9 Jul 2026.
Options
- A
+270 kJ
- B
−900 J
- C
+900kJ
- D
−900kJ
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