A) \[102.5litre\]
B) \[242.5litre\]
C) \[22.4\text{ }litre\]
D) \[121.25\text{ }litre\]
Correct Answer: A
Solution :
Since the gas is under S.T.P conditions, Hence \[{{V}_{1}}=22.4\] litre, \[{{V}_{2}}=?\] For isothermal and reversible expansion, \[\Delta E=0\] \[\therefore \] \[q=-W\] \[=-[-nRT\times 2.303\log \,\,\,{{V}_{2}}/{{V}_{1}}]\] or \[100\times 4.184\,J=1\times 8.314\times 273\] \[\times 2.303\times \log \frac{{{V}_{2}}}{22.4}\] or \[4184=5227.17\,\log \frac{{{V}_{2}}}{22.4}\] or \[\frac{{{V}_{2}}}{22.4}=\text{antilog}\,\,\text{0}\text{.8004}\] \[{{V}_{2}}=121.25ml\]You need to login to perform this action.
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