The pressure exerted by a non-reactive gaseou | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) First,calculate the number of moles of each gas:$n_{CH_4} = \frac{6.4 \ g}{16 \ g \ mol^{-1}} = 0.4 \ mol$$n_{CO_2} = \frac{8.8 \ g}{44 \ g \ mol^

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(D) First,calculate the number of moles of each gas:$n_{CH_4} = \frac{6.4 \ g}{16 \ g \ mol^{-1}} = 0.4 \ mol$$n_{CO_2} = \frac{8.8 \ g}{44 \ g \ mol^{-1}} = 0.2 \ mol$Total moles,$n = 0.4 + 0.2 = 0.6 \ mol$Convert volume to $m^3$:$V = 10 \ L = 10 	imes 10^{-3} \ m^3 = 0.01 \ m^3$Convert temperature to Kelvin:$T = 27 + 273 = 300 \ K$Using the ideal gas equation $PV = nRT$:$P = \frac{nRT}{V} = \frac{0.6 \ mol 	imes 8.314 \ J \ mol^{-1} K^{-1} 	imes 300 \ K}{0.01 \ m^3}$$P = 149652 \ Pa = 149.652 \ kPa$Rounding to the nearest integer,we get $150 \ kPa$.

Similar Questions

Which of the following equations does $NOT$ represent Charles's law for a given mass of gas at constant pressure?

For a fixed mass of an ideal gas,the following graph is given. Determine the correct relationship between $P_1, P_2$ and $P_3$.

What do the gas laws,such as Boyle's law,Charles's law,etc.,state?

$A$ gas occupies $0.418 \, L$ at $27 \, ^oC$ and $740 \, mm \, Hg$ pressure. If the mass of the gas is $3.0 \, g$,what is its molar mass?

Find the number of moles of $CO_2$ present in a sample occupying $4 \times 10^{-3} \text{ m}^3$ at $1.104 \times 10^5 \text{ Nm}^{-2}$ pressure. Given: $R \times T = 2208 \text{ J mol}^{-1}$. (in $\text{ mole}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module