A vessel of the volume \[50{\text{ litres}}\] contains an ideal gas at \[{0^ \circ }c\]. A portion of the gas is allowed to leak out from it under isothermal conditions so that pressure inside falls by \[0.8\] atmosphere. The number of moles of gas leaked out is nearly
(a)\[1.51{\text{ }}mole\]
(b) \[1.63{\text{ }}mole\]
(c) \[1.98{\text{ }}mole\]
(d) \[1.78{\text{ }}mole\]
Answer
524.7k+ views
Hint: When the temperature of an ideal gas remains constant, the isothermal process describes the relationship between volume and pressure. The volume of a gas is directly proportional to the number of moles of gas under constant temperature and pressure. By using the formula, the number of moles of gas can be found. The value of the universal gas constant has to be known to use the formula.
Complete step-by-step solution:
Given,
Volume, \[V = 50{\text{ litres}}\]
Temperature,\[T = {0^ \circ }c\]
For an ideal gas, we use the formula,
\[PV = nRT\]
Where,
\[P\]= pressure of the gas
\[V\]= volume of the gas
\[n\]= number of moles of gas
\[R\]= ideal gas constant and its value is
\[R = 0.08206{\text{ }}L.atm.mo{l^{ - 1}}{K^{ - 1}}\] and \[8.314{\text{ }}k.Pa.L.mo{l^{ - 1}}{K^{ - 1}}\]
and \[T\]= temperature of the gas at kelvin scale
initially,
\[PV = nRT\]
After the gas is leaked out final becomes,
\[P'V = n'RT \to \left( {P - P'} \right)V = \left( {n - n'} \right)RT\]
So,
No. of moles leaked out = \[\dfrac{{\left( {P - P'} \right)V}}{{RT}} = \dfrac{{0.8 \times 50}}{{0.0821 \times 273}} = 1.78moles\]
Note:When using the gas constant\[R = 8.31{\text{ }}J/K.mol\] , we must enter the pressure \[P\] in pascals\[Pa\] , volume in \[{m^3}\] , and temperature \[T\] in kelvin\[K\] .
When using the gas constant \[R = 0.08206{\text{ }}L.atm.mo{l^{ - 1}}{K^{ - 1}}\], pressure should be measured in atmospheres atm, volume measured in litres \[L\] , and temperature measured in kelvin \[K\].
When the particles of a gas are so far apart that they do not exert any attractive forces on one another, the gas is said to be ideal. There is no such thing as an ideal gas in real life, but at high temperatures and low pressures (conditions in which individual particles move very swiftly and are very far apart from one another, with essentially no interaction), gases behave very similarly so the Ideal Gas Law is a good approximation.
Complete step-by-step solution:
Given,
Volume, \[V = 50{\text{ litres}}\]
Temperature,\[T = {0^ \circ }c\]
For an ideal gas, we use the formula,
\[PV = nRT\]
Where,
\[P\]= pressure of the gas
\[V\]= volume of the gas
\[n\]= number of moles of gas
\[R\]= ideal gas constant and its value is
\[R = 0.08206{\text{ }}L.atm.mo{l^{ - 1}}{K^{ - 1}}\] and \[8.314{\text{ }}k.Pa.L.mo{l^{ - 1}}{K^{ - 1}}\]
and \[T\]= temperature of the gas at kelvin scale
initially,
\[PV = nRT\]
After the gas is leaked out final becomes,
\[P'V = n'RT \to \left( {P - P'} \right)V = \left( {n - n'} \right)RT\]
So,
No. of moles leaked out = \[\dfrac{{\left( {P - P'} \right)V}}{{RT}} = \dfrac{{0.8 \times 50}}{{0.0821 \times 273}} = 1.78moles\]
Note:When using the gas constant\[R = 8.31{\text{ }}J/K.mol\] , we must enter the pressure \[P\] in pascals\[Pa\] , volume in \[{m^3}\] , and temperature \[T\] in kelvin\[K\] .
When using the gas constant \[R = 0.08206{\text{ }}L.atm.mo{l^{ - 1}}{K^{ - 1}}\], pressure should be measured in atmospheres atm, volume measured in litres \[L\] , and temperature measured in kelvin \[K\].
When the particles of a gas are so far apart that they do not exert any attractive forces on one another, the gas is said to be ideal. There is no such thing as an ideal gas in real life, but at high temperatures and low pressures (conditions in which individual particles move very swiftly and are very far apart from one another, with essentially no interaction), gases behave very similarly so the Ideal Gas Law is a good approximation.
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