
A vessel of the volume \[8.3{\text{ }} \times {\text{ }}{10^{ - 3}}{m^3}\] contains an ideal gas at temperature \[27^\circ C\] and pressure $200\,kPa$. The gas is allowed to leak till the pressure falls to \[100{\text{ }}kPa\] and temperature increases to \[327^\circ C\]. The moles of gas leaked out is:
A) $\dfrac{2}{3}$ mole
B) $\dfrac{1}{3}$ mole
C) $\dfrac{1}{4}$ mole
D) $\dfrac{1}{2}$ mole
Answer
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Hint: In this solution, we will use the ideal gas law to determine the moles of gas inside the vessels. We will calculate the initial and the final moles to the gas to determine the moles of gas that leaked out.
Formula used: In this solution, we will use the following formula
Ideal gas law: $PV = nRT$ where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature of the gas.
Complete step by step answer:
We’ve been given a vessel containing a gas that leaks out. We know the initial pressure and volume and temperature as
${P_i} = 200 \times {10^3}Pa$, ${V_i} = 8.3 \times {10^{ - 3}}\,{m^3}$ and ${T_i} = 27^\circ C = 300\,K$
Then using the ideal gas law, we can write
$200 \times {10^3} \times 8.3 \times {10^{ - 3}}\,{m^3} = {n_i} \times 8.31 \times 300$
Now, we can determine the initial moles of the gas as
${n_i} = \dfrac{2}{3}\,{\text{moles}}$
After the gas leaks out, we have the final pressure, volume and temperature as
${P_f} = 100 \times {10^3}Pa$, ${V_f} = 8.3 \times {10^{ - 3}}\,{m^3}$ and ${T_f} = 327^\circ C = 600\,K$
Again using the ideal gas law, we can write
$100 \times {10^3} \times 8.3 \times {10^{ - 3}}\,{m^3} = {n_f} \times 8.31 \times 600$
Now, we can determine the final moles of the gas as
${n_f} = \dfrac{1}{6}\,{\text{moles}}$
Hence the change of moles of gas inside the vessel will be the difference between the final and the initial moles as
$\Delta n = {n_f} - {n_i}$
$ \Rightarrow \Delta n = \dfrac{1}{6} - \dfrac{2}{3} = - \dfrac{1}{2}$
Hence, the moles of gas leaked out is $\dfrac{1}{2}\,{\text{moles}}$ which corresponds to option (D).
Note: We must notice that since the gas is stored in the same vessel even in the final condition, the volume of the gas will still be equal to the volume of the vessel. The temperature of the gas should be taken in the Kelvin unit and not in the centigrade unit when using the ideal gas law.
Formula used: In this solution, we will use the following formula
Ideal gas law: $PV = nRT$ where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature of the gas.
Complete step by step answer:
We’ve been given a vessel containing a gas that leaks out. We know the initial pressure and volume and temperature as
${P_i} = 200 \times {10^3}Pa$, ${V_i} = 8.3 \times {10^{ - 3}}\,{m^3}$ and ${T_i} = 27^\circ C = 300\,K$
Then using the ideal gas law, we can write
$200 \times {10^3} \times 8.3 \times {10^{ - 3}}\,{m^3} = {n_i} \times 8.31 \times 300$
Now, we can determine the initial moles of the gas as
${n_i} = \dfrac{2}{3}\,{\text{moles}}$
After the gas leaks out, we have the final pressure, volume and temperature as
${P_f} = 100 \times {10^3}Pa$, ${V_f} = 8.3 \times {10^{ - 3}}\,{m^3}$ and ${T_f} = 327^\circ C = 600\,K$
Again using the ideal gas law, we can write
$100 \times {10^3} \times 8.3 \times {10^{ - 3}}\,{m^3} = {n_f} \times 8.31 \times 600$
Now, we can determine the final moles of the gas as
${n_f} = \dfrac{1}{6}\,{\text{moles}}$
Hence the change of moles of gas inside the vessel will be the difference between the final and the initial moles as
$\Delta n = {n_f} - {n_i}$
$ \Rightarrow \Delta n = \dfrac{1}{6} - \dfrac{2}{3} = - \dfrac{1}{2}$
Hence, the moles of gas leaked out is $\dfrac{1}{2}\,{\text{moles}}$ which corresponds to option (D).
Note: We must notice that since the gas is stored in the same vessel even in the final condition, the volume of the gas will still be equal to the volume of the vessel. The temperature of the gas should be taken in the Kelvin unit and not in the centigrade unit when using the ideal gas law.
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