The pressure exerted by 6.0 g of methane gas in a 0.03 m 3 vessel at 12 9 ∘ C is (Atomic masses: C = 12.01 , H = 1.01 and R = 8.314 JK − 1 mol − 1 )

w(given mass of methane) = 6 g temperature, T = 129 + 273 = 402 K mol mass of methane, M = 12.01 + 4 × 1.01 = 16.05 From, ideal gas equation, p V = n RT ⇒ P = v n RT p = 16.05 6 × 0.03 8.314 × 402 = 41648 Pa

Start Free Mock LoginAlready registered? Login here

Chemistry - Ideal Gas Question with Solution | TestHub

ChemistryIdeal GasIdeal Gas Equation & problemsMedium2 minPYQ_2012

ChemistryMediumsingle choice

The pressure exerted by $6.0 g$ of methane gas in $a 0.03 m^{3}$ vessel at $12 9^{\circ} C$ is (Atomic masses: $C = 12.01, H = 1.01$ and $R = 8.314 JK^{- 1} mol^{- 1}$ )

Options:

Answer:

Solution:

$w(given mass of methane) = 6 g temperature, T = 129 + 273 = 402 K mol mass of methane, M = 12.01 + 4 \times 1.01 = 16.05$ From, ideal gas equation, $p V = n RT \Rightarrow P = \frac{n RT}{v} p = \frac{6}{16.05} \times \frac{8.314 \times 402}{0.03} = 41648 Pa$

Stream:VITEEESubject:ChemistryTopic:Ideal GasSubtopic:Ideal Gas Equation & problems

⏱ 2mℹ️ Source: PYQ_2012

Doubts & Discussion

Loading discussions...