A vessel has 6g of oxygen at pressure P and temperature 400K. A small hole is made in it so that oxygen leaks out. How much oxygen leaks out if the final pressure is at $\dfrac{P}{2}$ and temperature is 300K?
$A.\;5g$
$B.\;4g$
$C.\;2g$
$D.\;3g$
Answer
547.3k+ views
Hint: - The ideal gas equation $PV = nRT$ provides a valuable model of the relations between volume, pressure, temperature, and the number of particles in a gas. Real molecules do have volume and do attract each other.
Complete step-by-step answer:
Formula Used: -
$PV = nRT$
Where $P$ is the pressure of the gas
$V$ is the volume
$n$ is the amount of substance or number of moles.
$R$ is ideal gas constant which is equal to $8.314\;{\text{Jmo}}{{\text{l}}^{ - 1}}{K^{ - 1}}$
$T$ is the temperature of the gas
$PV = \dfrac{m}{M}RT$, where $n = \dfrac{m}{M}$ which is $n$ amount of substance or number of moles, $m$ is total mass of gas in kilograms and $M$ is molar mass in kilogram per mole.
Now for, m grams of gas
$PV = \dfrac{m}{M}RT$----(i)
$P'V = \dfrac{{m'}}{M}RT'$---(ii)
In equation (i) and (ii) $P$is the initial pressure $P'$ is final pressure, similarly $m$ & $m'$ is the initial and final total mass of gas and $T$ & $T'$ is initial and final temperature of the gas.
Now we will divide equation (ii) by (i)
$\dfrac{{P'}}{P} = \dfrac{{m'T'}}{{mT}}$
$m' = \dfrac{{P'}}{P} \times \dfrac{T}{{T'}} \times m$
$\left( {\dfrac{\dfrac{P}{2}}{P}} \right)\times \dfrac{{400}}{{300}} \times 6 = 4g$
Since it is given in question that final pressure $P' = \dfrac{P}{2}$ and $T = 400\;K$ & $T' = 300\;K$
Therefore,
$\left( {\dfrac{1}{2}} \right) \times \dfrac{{400}}{{300}} \times 6 = 4g$
Mass of oxygen leaked, is equal to difference between initial and final total mass of gas which is given by:
$\Delta m = m - m' = 6 - 4 = 2g$
Hence option C is the correct answer and $2g$ is the amount of gas that leaks out.
Note: - For solving these types of questions we must remember the ideal gas equation which provides a valuable model of the relations between volume, pressure, temperature, and number of particles in a gas. As an ideal model, it serves as a reference for the behavior of real gases.
Complete step-by-step answer:
Formula Used: -
$PV = nRT$
Where $P$ is the pressure of the gas
$V$ is the volume
$n$ is the amount of substance or number of moles.
$R$ is ideal gas constant which is equal to $8.314\;{\text{Jmo}}{{\text{l}}^{ - 1}}{K^{ - 1}}$
$T$ is the temperature of the gas
$PV = \dfrac{m}{M}RT$, where $n = \dfrac{m}{M}$ which is $n$ amount of substance or number of moles, $m$ is total mass of gas in kilograms and $M$ is molar mass in kilogram per mole.
Now for, m grams of gas
$PV = \dfrac{m}{M}RT$----(i)
$P'V = \dfrac{{m'}}{M}RT'$---(ii)
In equation (i) and (ii) $P$is the initial pressure $P'$ is final pressure, similarly $m$ & $m'$ is the initial and final total mass of gas and $T$ & $T'$ is initial and final temperature of the gas.
Now we will divide equation (ii) by (i)
$\dfrac{{P'}}{P} = \dfrac{{m'T'}}{{mT}}$
$m' = \dfrac{{P'}}{P} \times \dfrac{T}{{T'}} \times m$
$\left( {\dfrac{\dfrac{P}{2}}{P}} \right)\times \dfrac{{400}}{{300}} \times 6 = 4g$
Since it is given in question that final pressure $P' = \dfrac{P}{2}$ and $T = 400\;K$ & $T' = 300\;K$
Therefore,
$\left( {\dfrac{1}{2}} \right) \times \dfrac{{400}}{{300}} \times 6 = 4g$
Mass of oxygen leaked, is equal to difference between initial and final total mass of gas which is given by:
$\Delta m = m - m' = 6 - 4 = 2g$
Hence option C is the correct answer and $2g$ is the amount of gas that leaks out.
Note: - For solving these types of questions we must remember the ideal gas equation which provides a valuable model of the relations between volume, pressure, temperature, and number of particles in a gas. As an ideal model, it serves as a reference for the behavior of real gases.
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