Real gases and Vander waal’s equation Questions in English

(B) The density of liquid water is $1 \,g/cc$. Therefore,$1 \,mol$ of water $(18 \,g)$ occupies an actual volume of $18 \,cc$.
At $STP$,$1 \,mol$ of water vapour occupies a total volume of $22400 \,cc$ $(22.4 \,L)$.
The percentage of volume actually occupied by the molecules is calculated as: $\frac{18 \,cc}{22400 \,cc} \times 100 \approx 0.08\%$.
Since $0.08\% < 1\%$,the correct option is $(B)$.

(B) The compressibility factor is defined as $Z = \frac{PV_m}{RT}$.
Given that $Z < 1$ at $STP$,where $P = 1 \ atm$ and $T = 273 \ K$.
For an ideal gas,$V_{ideal} = \frac{RT}{P} = \frac{0.0821 \times 273}{1} \approx 22.4 \ L$.
Since $Z = \frac{V_{real}}{V_{ideal}} < 1$,it follows that $V_{real} < V_{ideal}$.
Therefore,$V_m < 22.4 \ L$.

(D) The inversion temperature $(T_i)$ is given as $-80\,^{\circ}C$.
Converting this to Kelvin: $T_i = -80 + 273 = 193\,K$.
According to the Joule-Thomson effect,a gas produces cooling only when its temperature is below its inversion temperature $(T < T_i)$.
Among the given options,only $173\,K$ is less than $193\,K$.
Therefore,the gas will produce cooling at $173\,K$.

(B) The van der Waals equation is given by: $\left( P + \frac{n^2 a}{V^2} \right) (V - nb) = nRT$
Given: $n = 2 \, mol$,$V = 5 \, L$,$T = 27 + 273 = 300 \, K$,$a = 4.17$,$b = 0.03711$,$R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$.
Substituting the values: $\left( P + \frac{2^2 \times 4.17}{5^2} \right) (5 - 2 \times 0.03711) = 2 \times 0.0821 \times 300$
$\left( P + \frac{16.68}{25} \right) (5 - 0.07422) = 49.26$
$(P + 0.6672) \times 4.92578 = 49.26$
$P + 0.6672 = \frac{49.26}{4.92578} \approx 10.00$
$P = 10.00 - 0.6672 = 9.3328 \, atm \approx 9.33 \, atm$.

(B) The $Van \text{ der } Waals$ equation is given by $(P + \frac{an^2}{V^2})(V - nb) = nRT$. This equation is specifically derived to account for the deviations of real gases from ideal gas behaviour by considering molecular volume and intermolecular forces. Therefore,it explains the behaviour of $Real \text{ gases}$.

(B) Because molecules of real gases have intermolecular forces of attraction,the effective impact on the wall of the container is diminished.
The pressure of a real gas is reduced by a $\frac{a}{V^2}$ factor; hence,the behaviour of a real gas deviates from ideal behaviour.

(C) The Van der Waals equation for a non-ideal gas is given by $\left( P + \frac{a}{V^2} \right)(V - b) = RT$.
In this equation,the term $\left( P + \frac{a}{V^2} \right)$ accounts for the pressure correction due to intermolecular forces of attraction between gas molecules.
The term $\frac{a}{V^2}$ represents the reduction in pressure caused by these attractive forces,where $a$ is the Van der Waals constant representing the magnitude of intermolecular attraction.

(D) The Van der Waals equation for $1$ mole of a real gas is given by $\left( P + \frac{a}{V_m^2} \right) (V_m - b) = RT$,where $V_m$ is the molar volume $(V_m = V/n)$.
Substituting $V_m = V/n$ into the equation:
$\left( P + \frac{a}{(V/n)^2} \right) (V/n - b) = RT$
$\left( P + \frac{n^2 a}{V^2} \right) \left( \frac{V - nb}{n} \right) = RT$
Multiplying both sides by $n$ gives the expression for $n$ moles:
$\left( P + \frac{n^2 a}{V^2} \right) (V - nb) = nRT$

(C) Real gases behave like ideal gases at high temperature and low pressure.
Maximum deviation from ideal behavior is observed under conditions of low temperature and high pressure.
Comparing the given options,$-100\,^{\circ}C$ and $5 \ \text{atm}$ represents the lowest temperature and highest pressure combination,leading to the maximum deviation.

(D) The temperature at which the second virial coefficient of a real gas is zero is called Boyle's temperature.
It is the temperature at which a real gas obeys Boyle's law over an appreciable range of pressure.
It is given by the expression $T_{B} = \frac{a}{bR}$,where $a$ and $b$ are the van der Waals constants.
Boyle's temperature of a gas is always higher than its critical temperature.

(B) An ideal gas is defined as a system in which there are no intermolecular or interatomic forces.
Any real system approaches ideal gas behaviour in the limit where the pressure is extremely low and the temperature is high enough to overcome attractive intermolecular forces.
$PV = nRT$
Here,$n$ is the number of moles of gas.
The ideal gas equation describes the state of an ideal gas.
At low temperature and high pressure,gases deviate more from ideal behaviour because as pressure increases,the force of attraction between molecules increases and the molecules come closer together,making the volume of the gas particles significant compared to the total volume.

(D) In the Van der Waals equation,$(P + \frac{an^2}{V^2})(V - nb) = nRT$,the constant '$a$' represents the magnitude of the attractive forces between gas molecules.
The constant '$b$' represents the excluded volume or the effective volume occupied by the gas molecules.

(B) There are mainly $2$ reasons for the deviation of gases from ideal behavior: finite size of molecules and attractive forces between the molecules.
The finite size of molecules accounts for the factor '$b$' in the $Van \ der \ Waals$ equation,and the attractive forces between the molecules account for the factor '$a$' in the $Van \ der \ Waals$ equation.

(C) The Van der Waals equation is given by,
$(P + \frac{a}{V^2})(V - b) = RT$
At low pressure,the volume $V$ is very large,so the correction factor $b$ can be neglected compared to $V$.
At high temperature,the kinetic energy of molecules is high,making the attractive forces represented by the term $\frac{a}{V^2}$ negligible.
Under these conditions of low pressure and high temperature,the equation simplifies to $PV = RT$,which is the ideal gas equation.

(D) The ideal gas equation is $PV = nRT$.
Real gases approach ideal behavior under conditions of high temperature and low pressure,where the intermolecular forces are negligible and the volume of the gas particles is insignificant compared to the total volume of the container.

(C) An ideal gas is a hypothetical gas whose pressure,volume,and temperature behaviour is described by the ideal gas equation.
Real gases deviate from ideal behaviour due to the finite volume of gas particles and the existence of intermolecular forces.
Real gases approach ideal behaviour under conditions of low pressure and high temperature.
At low pressure,the volume of gas particles becomes negligible compared to the total volume.
At high temperature,the kinetic energy of the particles is high enough to overcome the intermolecular forces of attraction.
Comparing the given options,$0.5 \, atm$ (lowest pressure) and $500 \, K$ (highest temperature) provide the conditions where the real gas behaves most like an ideal gas.

(A) The Van der Waals equation for $1 \ mol$ of gas is: $(P + \frac{a}{V_m^2})(V_m - b) = RT$.
At low pressure,$V_m$ is very large,so $b$ is negligible compared to $V_m$ $(V_m - b \approx V_m)$.
The equation becomes: $(P + \frac{a}{V_m^2})V_m = RT$.
Expanding this: $PV_m + \frac{a}{V_m} = RT$.
Dividing by $RT$: $\frac{PV_m}{RT} + \frac{a}{V_m RT} = 1$.
Therefore,the compressibility factor $Z = \frac{PV_m}{RT} = 1 - \frac{a}{V_m RT}$.

(B) The Vander Waal's equation for $1$ mole of gas is $(p + \frac{a}{V_m^2})(V_m - b) = RT$.
At high temperature and low pressure,the volume of the gas $V_m$ is very large,so $V_m \gg b$,which means $(V_m - b) \approx V_m$.
Also,at low pressure,the term $\frac{a}{V_m^2}$ becomes negligible compared to $p$.
Substituting these approximations into the equation,we get $p V_m = RT$,which is the ideal gas equation.

(C) Helium behaves as a non-ideal gas.
When helium gas is allowed to expand into a vacuum,a heating effect is observed.
This occurs because the inversion temperature $(T_i)$ of helium is very low (approximately $-240 \ ^\circ C$ or $33 \ K$).
For most gases,the Joule-Thomson expansion results in a cooling effect because the work done during adiabatic expansion is at the cost of internal energy.
However,for gases like $H_2$ and $He$,the temperature of the gas is above their respective inversion temperatures at room temperature,leading to a heating effect upon expansion.

(A) The correct option is $A$. In $Vander$ $Waal$'s equation,$(P + \frac{an^2}{V^2})(V - nb) = nRT$,the constant '$b$' is known as the co-volume or excluded volume. It represents the effective volume occupied by the gas molecules themselves.

(A) The van der Waals equation is $(P + \frac{n^2a}{V^2})(V - nb) = nRT$.
Given: $n = 1 \, mol$,$V = 0.25 \, L$,$T = 300 \, K$,$a = 2.253 \, atm \, L^2 \, mol^{-2}$,$b = 0.0428 \, L \, mol^{-1}$,and $R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$.
Substituting the values: $(P + \frac{1^2 \times 2.253}{0.25^2})(0.25 - 1 \times 0.0428) = 1 \times 0.0821 \times 300$.
$(P + \frac{2.253}{0.0625})(0.2072) = 24.63$.
$(P + 36.048)(0.2072) = 24.63$.
$P + 36.048 = \frac{24.63}{0.2072} \approx 118.87$.
$P = 118.87 - 36.048 = 82.822 \, atm$.

(D) The critical temperature $(T_c)$ of a gas is given by the formula: $T_c = \frac{8a}{27Rb}$.
From the formula,$T_c$ is directly proportional to $a$ and inversely proportional to $b$ $(T_c \propto \frac{a}{b})$.
Calculating the ratio $\frac{a}{b}$ for each gas:
For $W$: $\frac{4.0}{0.027} \approx 148.15$
For $X$: $\frac{8.0}{0.030} \approx 266.67$
For $Y$: $\frac{6.0}{0.032} = 187.5$
For $Z$: $\frac{12.0}{0.027} \approx 444.44$
Since gas $Z$ has the highest ratio of $\frac{a}{b}$,it has the highest critical temperature.

(C) The Van der Waals constant '$a$' is a measure of the magnitude of the attractive forces between the gas molecules.
Greater the value of '$a$',stronger are the intermolecular forces of attraction.
Stronger intermolecular forces make it easier for a gas to be liquefied.
Comparing the given values: $a(NH_3) = 4.170 > a(CH_4) = 2.253 > a(N_2) = 1.390 > a(O_2) = 1.3$.
Since $NH_3$ has the highest value of '$a$',it can be most easily liquefied.

(C) Real gases deviate from ideal gas behavior under conditions of low temperature and high pressure.
Comparing the given options,the condition with the lowest temperature $(-100 \ ^oC)$ and the highest pressure $(4 \ atm)$ will result in the greatest deviation from ideal gas behavior.
Therefore,the correct option is $C$.

(B) The compressibility factor $Z$ is defined as $Z = \frac{V_{m, \text{real}}}{V_{m, \text{ideal}}}$.
Given that $Z < 1$,it implies that $V_{m, \text{real}} < V_{m, \text{ideal}}$.
At $STP$,the molar volume of an ideal gas $V_{m, \text{ideal}}$ is $22.4 \ L$.
Therefore,for a gas with $Z < 1$,the molar volume $V_m$ will be less than $22.4 \ L$.

(C) The cooling in a refrigerator is based on the $Joule-Thomson$ effect.
When a compressed gas is allowed to expand suddenly through a nozzle or a small orifice,it undergoes cooling.
This principle is utilized in the cooling cycle of a refrigerator.

(D) In the Van der Waals equation,the constant '$a$' represents the magnitude of the attractive forces between gas molecules,while the constant '$b$' represents the excluded volume occupied by the gas molecules.

(C) The $van \ der \ Waals$ equation for $n$ moles of a gas is given by: $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
For $1$ mole of gas,it is $(P + \frac{a}{V_m^2})(V_m - b) = RT$.
At low pressure,the volume $V_m$ is very large,so $b$ becomes negligible compared to $V_m$.
At high temperature,the pressure $P$ is significant compared to the internal pressure term $\frac{a}{V_m^2}$,making $\frac{a}{V_m^2}$ negligible.
Thus,the equation reduces to $PV_m = RT$,which is the ideal gas equation.

(A) The van der Waals equation for $1 \ mol$ of a gas is given by $(p + \frac{a}{V_m^2})(V_m - b) = RT$.
At high temperature,the kinetic energy of gas molecules is very high,making the effect of intermolecular forces $(a/V_m^2)$ negligible.
At low pressure,the volume of the gas $(V_m)$ is very large,making the excluded volume $(b)$ negligible compared to $V_m$.
Under these conditions,the equation reduces to $pV_m = RT$,which is the ideal gas equation.

(A) real gas behaves like an ideal gas when the intermolecular forces are negligible and the volume occupied by the gas molecules is negligible compared to the total volume of the gas.
This occurs at $High \ Temperature$ and $Low \ Pressure$.

(C) The van der Waals equation for $n$ moles of a real gas is given by: $\left( P + \frac{an^2}{V^2} \right)(V - nb) = nRT$.
For $1$ mole of gas $(n=1)$,the equation becomes: $\left( P + \frac{a}{V^2} \right)(V - b) = RT$.
In this equation,the term $\frac{a}{V^2}$ accounts for the intermolecular forces of attraction between gas molecules.
Therefore,the term representing intermolecular forces is $\left( P + \frac{a}{V^2} \right)$.

(D) The van der Waals equation for $n$ moles of a real gas is derived by correcting the pressure and volume terms of the ideal gas equation.
Pressure correction: $P_{ideal} = P_{real} + \frac{an^2}{V^2}$
Volume correction: $V_{ideal} = V_{real} - nb$
Substituting these into the ideal gas equation $PV = nRT$,we get:
$\left( {P + \frac{{{n^2}a}}{{{V^2}}}} \right)(V - nb) = nRT$

(A) The Van der Waals equation for real gases is given by: $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
From the term $\frac{an^2}{V^2}$,which is added to pressure $(P)$,the units of $\frac{an^2}{V^2}$ must be the same as the units of pressure $(atm)$.
Therefore,$unit(a) = unit(P) \times \frac{unit(V)^2}{unit(n)^2}$.
Substituting the units: $unit(a) = atm \times \frac{L^2}{mol^2} = atm\,L^2\,mol^{-2}$.

(C) The van der Waals equation for $n$ moles of a gas is: $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
For $1$ mole of gas $(n=1)$,the equation becomes: $(P + \frac{a}{V^2})(V - b) = RT$.
Expanding this,we get: $PV - Pb + \frac{a}{V} - \frac{ab}{V^2} = RT$.
At high pressure,the volume $V$ is very small,so the term $\frac{a}{V}$ is negligible compared to $P$. Also,the term $\frac{ab}{V^2}$ is very small and can be neglected.
Thus,the equation simplifies to: $PV - Pb = RT$,which rearranges to $PV = RT + Pb$.

(B) The compressibility factor is defined as $Z = \frac{PV_m}{RT}$.
At $S.T.P.$,$P = 1 \, bar$ and $T = 273.15 \, K$,the molar volume of an ideal gas is $V_{ideal} = 22.4 \, L$.
Given $Z < 1$,we have $\frac{PV_m}{RT} < 1$,which implies $V_m < \frac{RT}{P}$.
Since $\frac{RT}{P} = 22.4 \, L$ at $S.T.P.$,it follows that $V_m < 22.4 \, L$.

(C) Real gases deviate from ideal behavior due to intermolecular forces and the finite volume of gas molecules.
At $High \ temperature$,the kinetic energy of molecules is high,making intermolecular forces negligible.
At $Low \ pressure$,the volume occupied by the gas molecules becomes negligible compared to the total volume of the container.
Therefore,a real gas approaches ideal behavior at $High \ temperature$ and $Low \ pressure$.

(B) Real gases deviate from ideal behavior because of intermolecular forces and the finite volume of gas molecules.
These deviations become significant when the gas molecules are close together and moving slowly.
This occurs at $Low \ temperature$ (where kinetic energy is low) and $High \ pressure$ (where molecules are compressed into a smaller volume).
Therefore,the deviation from the ideal gas equation $PV = nRT$ is maximum at $Low \ temperature$ and $High \ pressure$.

(D) In an ideal gas,it is assumed that there are no intermolecular forces of attraction between the gas molecules.
However,in real gases,there exist intermolecular forces of attraction.
When a molecule approaches the wall of the container to exert pressure,it is pulled back by the other molecules in the bulk due to these attractive forces.
As a result,the molecule hits the wall with less force than it would have in the absence of these attractions.
Therefore,the observed pressure of a real gas is lower than the pressure of an ideal gas.

(C) In the $Vander$ $Waals$ equation,$(P + \frac{an^2}{V^2})(V - nb) = nRT$,the constant $'b'$ is known as the co-volume or excluded volume.
It represents the effective volume occupied by the gas molecules themselves.

(C) The van der Waals constant $a$ represents the magnitude of intermolecular attractive forces between gas molecules.
Greater the value of $a$,stronger are the intermolecular forces of attraction.
Stronger intermolecular forces facilitate easier liquefaction of the gas.
Comparing the given values: $a(NH_3) = 4.170 > a(CH_4) = 2.253 > a(N_2) = 1.390 > a(O_2) = 1.360$.
Therefore,$NH_3$ has the highest value of $a$ and can be liquefied most easily.

(B) The van der Waals equation for $n$ moles of a real gas is given by: $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
In this equation,the term $\frac{an^2}{V^2}$ represents the pressure correction due to the intermolecular attractive forces between gas molecules.
Therefore,the term indicating the presence of attractive forces is $\frac{an^2}{V^2}$.

(D) The compressibility factor $Z$ is defined as $Z = \frac{PV}{nRT}$.
For a van der Waals gas,the equation is $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
At low pressure,$Z \approx 1 - \frac{a}{RTV}P$,which explains the initial decrease in $Z$ for real gases (where $a > 0$).
At high pressure,$Z \approx 1 + \frac{Pb}{RT}$,which explains the increase in $Z$ for real gases (where $b > 0$).
Gas $A$ shows $Z > 1$ for all $P$,implying $a = 0$ and $b > 0$.
Gas $B$ shows $Z < 1$ for all $P$,implying $a > 0$ and $b = 0$.
The dashed curve represents a typical real gas where both $a$ and $b$ are non-zero.
Therefore,the statement that the dashed curve represents an ideal gas is false.

(D) real gas behaves like an ideal gas under conditions of $High \ temperature$ and $Low \ pressure$.
At high temperatures,the kinetic energy of the gas molecules is high,making the intermolecular forces of attraction negligible.
At low pressures,the volume occupied by the gas molecules is negligible compared to the total volume of the container,satisfying the assumptions of the kinetic molecular theory of gases.

(C) The Van der Waals constant $a$ represents the magnitude of intermolecular attractive forces.
Larger and more complex molecules have stronger intermolecular forces,leading to higher $a$ values.
Comparing the given gases:
$1. C_6H_6$ (Benzene) has $a \approx 18.000$.
$2. C_6H_5CH_3$ (Toluene) is larger than benzene,so it has the highest $a \approx 24.060$.
$3. Ne$ (Neon) is a noble gas with weak dispersion forces,so it has the lowest $a \approx 0.217$.
$4. H_2O$ (Water) has hydrogen bonding,giving $a \approx 5.464$.
Thus,the correct matching is: $1-c, 2-d, 3-a, 4-b$.

(C) Real gases behave most like ideal gases at low pressure and high temperature.
Comparing the given conditions,the condition with the lowest pressure $(0.5 \ atm)$ and the highest temperature $(500 \ K)$ will show the most ideal behavior.

(B) The van der Waals equation is valid for $Real \ gases$.
It is essentially a modified form of the ideal gas equation,$(P + \frac{an^2}{V^2})(V - nb) = nRT$,which accounts for intermolecular forces and the finite volume of gas molecules.

(B) The constant $ 'a'$ in the van der Waals equation represents the magnitude of intermolecular attractive forces.
Greater the value of $ 'a'$,stronger are the intermolecular forces,which makes the gas easier to liquefy.
Conversely,a smaller value of $ 'a'$ indicates weaker intermolecular forces,making the gas more difficult to liquefy.
Comparing the given values: $N_2 (1.35) < O_2 (1.36) < CO (1.46) < CH_4 (2.25)$.
Since $N_2$ has the lowest value of $ 'a'$,it has the weakest intermolecular forces and will be the most difficult to liquefy.

(C) The compressibility factor $Z$ is defined as $Z = \frac{pV}{nRT}$.
When $Z > 1$,the repulsive forces dominate,making the gas less compressible than an ideal gas.
Therefore,it is difficult to compress the gas compared to an ideal gas.