A) 212 kPa
B) 209 kPa
C) 206 kPa
D) 200 kPa
Correct Answer: B
Solution :
[b] \[\frac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}\] Here, \[{{P}_{1}}=200kPa\] \[{{T}_{1}}=22{}^\circ C=295K~~~~~~{{T}_{2}}=42{}^\circ C=315K\] \[{{V}_{2}}={{V}_{1}}+\frac{2}{100}{{V}_{1}}=1.02{{V}_{1}}\] \[\therefore \,\,\,\,\,{{P}_{2}}=\frac{200\times 315{{V}_{1}}}{295\times 1.02{{V}_{1}}}=209.97kPa\]You need to login to perform this action.
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