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Real Gases

1.25 moles of...

Question

1.25 moles of a gas A and y moles of a gas B exert a pressure of 300 Pa in a container of volume $10 m^{3}$ at $1000 K$ . Find the value of y in terms of R (where R is the gas constant in $J K^{- 1} m o l^{- 1}$ )

\frac{12}{5 R}

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\frac{3}{R}

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\frac{12 - 5 R}{4 R}

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\frac{15 - 4 R}{5 R}

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Solution

The correct option is C $\frac{12 - 5 R}{4 R}$
Total number of moles in the container (n) = 1.25 + y
Pressure = 300 Pa
Volume= 10 L
Temperature = 1000 K
We know that
$P V = n R T$
$300 \times 10 = (1.25 + y) R \times 1000 \Rightarrow y = \frac{12 - 5 R}{4 R}$

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1.25 moles of a gas A and y moles of a gas B exert a pressure of 300 Pa in a container of volume 10 m3 at 1000 K. Find the value of y in terms of R (where R is the gas constant in JK−1mol−1 )

The correct option is C 12−5R4RTotal number of moles in the container (n) = 1.25 + y Pressure = 300 Pa Volume= 10 L Temperature = 1000 K We know that PV = nRT 300×10=(1.25+y)R×1000 ⇒y=12−5R4R

1.25 moles of a gas A and y moles of a gas B exert a pressure of 300 Pa in a container of volume $10 m^{3}$ at $1000 K$ . Find the value of y in terms of R (where R is the gas constant in $J K^{- 1} m o l^{- 1}$ )

The correct option is C $\frac{12 - 5 R}{4 R}$
Total number of moles in the container (n) = 1.25 + y
Pressure = 300 Pa
Volume= 10 L
Temperature = 1000 K
We know that
$P V = n R T$
$300 \times 10 = (1.25 + y) R \times 1000 \Rightarrow y = \frac{12 - 5 R}{4 R}$