A) 212 kPa
B) 209 kPa
C) 206 kPa
D) 200 kPa
Correct Answer: B
Solution :
By ideal gas equation \[\operatorname{pV} = nRT\] \[\frac{{{p}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{p}_{2}}{{V}_{2}}}{{{T}_{2}}}\] Given, \[{{\operatorname{p}}_{1}} = 200 kPa,\,\,{{V}_{1}}=V\] \[{{T}_{1}}= 273+22\] \[=\,\,\,295\,\,K\] \[{{V}_{2}} =V +0.02 V\] \[{{T}_{2}} =273+42= 315 K\] \[\therefore \,\,\,\,\,\,\frac{200\times V}{295}=\frac{{{p}_{2}}\times 1.02\,V}{315}\] \[{{p}_{2}}=\frac{200\times 315}{295\times 1.02}=209\,\,kPa\]You need to login to perform this action.
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