Question

The pressure exerted by 6.0g of methane gas in a 0.03 m³
vessel at 129°C is (Atomic masses : C = 12.01, H = 1.01 and
R = 8.314 Kpa dm³K⁻¹ mol⁻¹)
(a) 31684 Pa (b) 215216 Pa
(c) 13409 Pa (d) 41777 Pa

Answer 1

Dear Student,

◆ Answer -

(d) 41777 Pa

● Explanation -

# Given -

V = 0.03 m³

T = 129 °C = 402 K

W = 6 g

# Solution -

Molar mass of methane is

M = 12 + 4×1

M = 16 g

Let P be the pressure exerted by methane.

PV = nRT

PV = WRT/M

P × 0.03 = 6 × 8.314 × 402 / 16

P = 6 × 8.314 × 402 / (16 × 0.03)

P = 41777 Pa

Hence, be the pressure exerted by methane on vessel is 41777 Pa.

Thanks dear. Hope this helps you..

Answer 2

The excreted pressure is 41777 pa.

Correct answer - "d".

Step by step explanation:

From the given,

Mass of methane = w = 6.0 g

Volume of methane = V = $0.03m^{3}$

Temperature = T = $129+273=402K$

R = $8.314\,JK^{-1}mol^{-1}$

Molecular mass of methane = $CH_{4}$ = $12.01+4(1.01)=16.05$.

According to the ideal gas law,

                            $\bold{Pv =nRT=\frac{w}{M}RT}$.................(1)

We have to find the "P" in the above expression.

Rearrange the equation "1" is as follows.

                                  $\bold{P=\frac{w}{M}\frac{RT}{V}}$............................(2)

Substitute the all given values in the equation (2).

                                      $\Rightarrow \frac{6}{16.05}\times \frac{8.314\times 402}{0.03}$

                                     $\Rightarrow 41777Pa$

Therefore, The excreted pressure is 41777 pa.

Hence, Correct answer - "d".

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