1 L closed flask contains a mixture of 4 g | vedclass

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

Question

(A) Step $1$: Calculate the number of moles of each gas.$n_{CH_4} = \frac{4 \ g}{16 \ g/mol} = 0.25 \ mol$$n_{CO_2} = \frac{4.4 \ g}{44 \ g/mol} = 0.1

Vedclass · Accepted Answer

$8.6$

Vedclass · Answer

(A) Step $1$: Calculate the number of moles of each gas.$n_{CH_4} = \frac{4 \ g}{16 \ g/mol} = 0.25 \ mol$$n_{CO_2} = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$Step $2$: Calculate the total number of moles $(n_T)$.$n_T = 0.25 \ mol + 0.1 \ mol = 0.35 \ mol$Step $3$: Use the ideal gas equation $PV = nRT$ to find the pressure.Given: $V = 1 \ L$, $T = 27 + 273 = 300 \ K$, $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.$P = \frac{n_T RT}{V} = \frac{0.35 	imes 0.082 	imes 300}{1} = 8.61 \ atm \approx 8.6 \ atm$.

$1 \ L$ closed flask contains a mixture of $4 \ g$ of methane and $4.4 \ g$ of carbon dioxide. The pressure inside the flask at $27^{\circ} C$ is [Assume ideal behaviour of gases] (in $atm$)

Similar Questions

What is the molar volume of an ideal gas at $27\,^oC$ and $1\, atm$ pressure in $L$?

When does $Z = 1$ occur?

$A$ certain mass of a gas occupies a volume of $250 \ mL$ at $2 \ atm$ pressure. Calculate the volume of the gas if the pressure is increased to $2.5 \ atm$ at a constant temperature. (in $mL$)

In the equation $PV = nRT$,which one cannot be the numerical value of $R$?

$4 \ g$ of an ideal gas occupies $5.6035 \ L$ of volume at $546 \ K$ and $2 \ atm$ pressure. What is its molecular weight?

Vedclass Test Series

Exam Paper Generator

Online Exam Module