A) 215216 Pa
B) 13409 Pa
C) 41648 Pa
D) 31684 Pa
Correct Answer: C
Solution :
Given, volume, \[V=0.03\,\,{{m}^{3}}\] temperature, T = 129 + 273 = 402 K mass of methane, w = 6.0 g mol. mass of methane, \[M=12.01+4\times 1.01=16.05\] From, ideal gas equation, \[pV=nRT\] \[p=\frac{6}{16.05}\times \frac{8.314\times 402}{0.03}=41648Pa\]You need to login to perform this action.
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