A) 0.01 kg
B) 0.001 kg
C) 0.002 kg
D) 0.003 kg
Correct Answer: B
Solution :
Gas equation is, \[pV=nRT\] or \[pV=\frac{m}{M}RT\] \[3.53\times {{10}^{5}}\times 1\times {{10}^{-3}}\] \[=\frac{m}{32\times {{10}^{-3}}}\times R\times 320\] ...(i) On solving, \[m=\frac{3.53}{R\times {{10}^{2}}}kg\] Let\[x\]kg mass of oxygen is used at constant temperature, so \[pV=\frac{m-x}{M}RT\] \[2.7\times {{10}^{5}}\times 1\times {{10}^{-3}}=\frac{m-x}{32\times {{10}^{-3}}}R\times 320\]...(ii) \[m-x=\frac{2.70}{R\times {{10}^{2}}}kg\] From Eqs. (i) and (ii) \[\frac{m-x}{m}=\frac{270}{353}\] \[\Rightarrow \] \[x=\frac{83}{353}m\] \[=\frac{83}{353}.\frac{3.53}{R\times {{10}^{2}}}\] \[=\frac{83}{353}\times \frac{3.53}{8.3\times {{10}^{2}}}\] \[=0.001\text{ }kg\]You need to login to perform this action.
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