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Question

Calculate the maximum work done (in cal) when the pressure on 10 grams of hydrogen is reduced from 20 to 1 atm at a constant temperature of 273 K if the gas behaves ideally.
  1. +8180 cal
  2. 8810 cal
  3. 8180 cal
  4. +8810 cal


Solution

The correct option is C 8180 cal
Maximum work done is obtained when the process is reversible.
So, work done in an isothermal reversible process is given as:
w=2.303nRT logP1P2

n = number of moles of hydrogen 
n=given weightmolar mass = 102=5 mol

Given : P1=20 atm
P2=1 atm
T=273 K
R=2 cal/K.mol
So,
w=2.303×5×2×273×log201
= 8179.8 Cal which is about 8180 Cal

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