| The pressure exerted by 6.0 g of methane gas in a \[0.03\text{ }{{m}^{3}}\] vessel at \[129{}^\circ C\] is (Atomic masses: [AIPMT (M) 2010] |
| \[C=12.01,H=1.01\,\] and \[R=8.314J{{K}^{-1}}\,mo{{l}^{-1}}\] |
A) 215216 Pa
B) 13409 Pa
C) 41648 Pa
D) 31684 Pa
Correct Answer: C
Solution :
| Given, |
| volume, V = 0,03 m3 temperature, T = 129 + 273 = 402 K mass of methane, w = 6.0 g mol. mass of methane, M = 12.01 + 4 x 1.01 |
| = 16.05 |
| From, ideal gas equation, |
| \[pV=nRT\] |
| \[p=\frac{6}{16.05}\times \frac{8.314\times 402}{0.03}=41648Pa\] |
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