1
The pressure exerted by 6.0 g of methane gas in a 0.03 m3 vessel at 129 is (Atomic masses : C = 12.01, H = 1.01 and R = 8.314 J K-1 mol-1)
- 215216 Pa
- 13409 Pa
- 41648 Pa
- 31684 Pa
Answer
(3) 41648 Pa
Solution
Molar mass of CH4 = 12.01 + 4 × 1.01 = 16.05 g mol-1
n = 6.016.05 = 0.3738 mol
T = 129 + 273 = 402 K
pV = nRT
p = nRTV
= 0.3738 8.314 4020.03
= 41648 Pa
Answer: (3)
Out of syllabusAIPMT 2010 (Mains)