Real gases and Vander waal’s equation Questions in English

(C) Real gases deviate from ideal behavior due to intermolecular forces and the finite volume of gas molecules.
According to the kinetic molecular theory,real gases approach ideal behavior under conditions where intermolecular forces are negligible and the volume of particles is insignificant compared to the total volume.
These conditions are achieved at $low$ pressure and $high$ temperature.
Therefore,option $C$ is correct.

(A) The compressibility factor $Z$ is defined by the formula $Z = \frac{PV}{nRT}$.
Given values are $P = 50 \ atm$,$V = 0.35 \ L$,$n = 1 \ mol$,$T = 300 \ K$,and the gas constant $R = 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.
Substituting these values into the formula:
$Z = \frac{50 \times 0.35}{1 \times 0.0821 \times 300}$
$Z = \frac{17.5}{24.63}$
$Z \approx 0.71$.

(A) In the ideal gas equation,we neglect the intermolecular attractions and the volume occupied by molecules. However,the real gas equation accounts for both of these. The parameter $a$ is a measure of intermolecular attraction,and the parameter $b$ is a measure of the volume occupied by the molecules. In particular,if the volume of a single gas molecule is $v$,then $b = 4vN_A$. Hence,$b$ is four times the molar volume of the gas molecules.

(D) Real gases deviate from ideal behavior because of intermolecular forces and the finite volume of gas molecules.
This deviation is most significant under conditions of low temperature and high pressure.
Comparing the given options,the condition with the lowest temperature $(-13\,^{\circ}C)$ and the highest pressure $(2\, atm)$ will show the maximum deviation from ideal behavior.
Therefore,the correct option is $D$.

(C) The van der Waals equation for $1$ mole of gas is given by: $\left(P + \frac{a}{V^2}\right)(V - b) = RT$.
At relatively high pressure,the volume $V$ is small,making the pressure correction term $\frac{a}{V^2}$ negligible compared to $P$.
However,the volume correction term $b$ remains significant.
Thus,the equation simplifies to: $P(V - b) = RT$.
Expanding this,we get: $PV - Pb = RT$,which rearranges to $PV = RT + Pb$.
Therefore,option $C$ is correct.

(A) The given equation is $(P + \frac{a}{V^2})V = RT$.
Expanding the equation: $PV + \frac{a}{V} = RT$.
Dividing both sides by $RT$: $\frac{PV}{RT} + \frac{a}{RTV} = 1$.
Since the compressibility factor $Z = \frac{PV}{RT}$,we substitute $Z$ into the equation: $Z + \frac{a}{RTV} = 1$.
Rearranging for $Z$: $Z = 1 - \frac{a}{RTV}$.

(C) The deviation from ideal behaviour is expressed by the compressibility factor $Z.$
$Z$ is defined as $Z = \frac{PV}{nRT}.$ For an ideal gas,$PV = nRT,$ so $Z = 1.$
When $Z > 1,$ the gas shows positive deviation from ideal behaviour.
This implies that the gas is less compressible than an ideal gas,meaning real gases are difficult to compress compared to ideal gases.

(C) The van der Waals equation for $1$ mole of gas is $\left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT$.
At high pressure,the molar volume $V_m$ is small,making the term $\frac{a}{V_m^2}$ negligible compared to $P$.
Thus,the equation simplifies to $P(V_m - b) = RT$.
Expanding this gives $PV_m - Pb = RT$,which rearranges to $PV_m = RT + Pb$.
Therefore,option $C$ is correct.

(B) Real gases behave ideally under conditions of low pressure and high volume,as the intermolecular forces become negligible and the volume of gas molecules becomes insignificant compared to the total volume of the container.
At $Boyle's$ temperature,the effects of intermolecular forces and molecular volume cancel out,leading to ideal behavior.
However,at very low temperatures,the kinetic energy of gas molecules decreases,making intermolecular forces significant,which causes real gases to deviate from ideal behavior.
Therefore,the statement that real gases show ideal behavior at very low temperature is incorrect.

(A) For one mole of a real gas,the van der Waals equation is given by:
$(P + \frac{a}{V^2})(V - b) = RT$
At very high pressures,the volume $V$ is small,so the term $\frac{a}{V^2}$ becomes negligible compared to $P$.
Thus,the equation simplifies to:
$P(V - b) = RT$
$PV - Pb = RT$
$PV = RT + Pb$
Dividing both sides by $RT$:
$\frac{PV}{RT} = 1 + \frac{Pb}{RT}$
Since the compressibility factor $Z = \frac{PV}{RT}$,we get:
$Z = 1 + \frac{Pb}{RT}$

(C) gas deviates from ideal behaviour when the assumptions of the kinetic molecular theory are no longer valid.
At $high$ pressure,the volume occupied by the gas molecules becomes significant compared to the total volume of the container.
At $low$ temperature,the kinetic energy of the molecules decreases,making intermolecular forces of attraction significant.
Therefore,at $high$ pressure and $low$ temperature,the gas deviates the most from its ideal behaviour.

(A) The Van der Waals equation is given by $\left( P + \frac{n^2a}{V^2} \right)(V - nb) = nRT$.
At low pressure,the volume $V$ is very large,making the correction term $\frac{n^2a}{V^2}$ negligible compared to $P$.
At high temperature,the volume $V$ is also large,making the correction term $nb$ negligible compared to $V$.
Under these conditions,the equation simplifies to $PV = nRT$,which is the ideal gas equation.

(B) The compressibility factor $Z$ is defined as $Z = \frac{PV}{nRT}$.
Rearranging for pressure,we get $P = \frac{ZnRT}{V}$.
At constant temperature $T$ and equal number of moles $n$,the pressure $P$ is directly proportional to $\frac{Z}{V}$,i.e.,$P \propto \frac{Z}{V}$.
Given $V_A = 2V_B$ and $Z_A = 3Z_B$.
Taking the ratio of pressures:
$\frac{P_A}{P_B} = \frac{Z_A}{Z_B} \times \frac{V_B}{V_A} = \left( \frac{3}{1} \right) \times \left( \frac{1}{2} \right) = \frac{3}{2}$.
Therefore,$2P_A = 3P_B$.

(C) The critical temperature $(T_c)$ of a gas is given by the formula $T_c = \frac{8a}{27Rb}$.
To find the gas with the highest $T_c$,we need to calculate the ratio $\frac{a}{b}$ for each gas:
$Ar: \frac{1.3}{3.2} \approx 0.406$
$Ne: \frac{0.2}{1.7} \approx 0.118$
$Kr: \frac{5.1}{1.0} = 5.1$
$Xe: \frac{4.1}{5.0} = 0.82$
Since $Kr$ has the highest $\frac{a}{b}$ ratio,it will have the highest critical temperature.

(C) The given equation of state is $P(V - b) = RT$.
Expanding this,we get $PV - Pb = RT$.
Dividing by $RT$,we obtain $\frac{PV}{RT} - \frac{Pb}{RT} = 1$.
Since the compression factor $Z = \frac{PV}{RT}$,the equation becomes $Z = 1 + \frac{Pb}{RT}$.
This shows that $Z$ increases linearly with $P$ with a slope of $\frac{b}{RT}$.
Therefore,the gas with the largest van der Waals constant $b$ will exhibit the steepest increase in the plot of $Z$ versus $P$.
The magnitude of $b$ depends on the size of the gas molecules,which follows the order $Ne < Ar < Kr < Xe$.
Thus,$Xe$ has the largest value of $b$ and will show the steepest increase.

(D) The constant $a$ represents the magnitude of intermolecular forces of attraction. $A$ higher value of $a$ implies stronger attraction,which pulls molecules closer,resulting in a smaller volume occupied.
For gases $A$ and $C$,$b$ is the same $(0.05196)$. Since $a_A (642.32) > a_C (431.91)$,gas $A$ has stronger intermolecular forces and occupies less volume than gas $C$. Thus,gas $C$ occupies more volume than gas $A$.
For gases $B$ and $D$,$a$ is the same $(155.21)$. The constant $b$ represents the excluded volume per mole. $A$ smaller value of $b$ means the molecules occupy less space,making the gas more compressible.
Since $b_B (0.04136) < b_D (0.4382)$,gas $B$ is more compressible than gas $D$.

(B) The van der Waals gas equation for $1 \ mol$ of gas is $(P + \frac{a}{V^2})(V - b) = RT$.
At low pressure,the volume $V$ is very large,so $V - b \approx V$.
Substituting this into the equation: $(P + \frac{a}{V^2})V = RT$.
Expanding the terms: $PV + \frac{a}{V} = RT$.
Rearranging gives: $PV = RT - \frac{a}{V}$.

(B) The van der Waals equation for a real gas is $(P + \frac{a}{V^2})(V - b) = RT$.
At high pressure,the term $\frac{a}{V^2}$ is negligible compared to $P$.
Thus,the equation simplifies to $P(V - b) = RT$.
Expanding this,we get $PV - Pb = RT$.
Dividing both sides by $RT$,we get $\frac{PV}{RT} - \frac{Pb}{RT} = 1$.
Since the compressibility factor $Z = \frac{PV}{RT}$,we have $Z - \frac{Pb}{RT} = 1$.
Therefore,$Z = 1 + \frac{Pb}{RT}$.
Hence,the correct option is $B$.

(B) For an ideal gas,$PV = nRT$. Given $n = 1 \, mol$,$V = 22.4 \, L$,and $T = 273 \, K$,the ideal pressure $P_{ideal} = \frac{nRT}{V} = \frac{1 \times 0.0821 \times 273}{22.4} \approx 1 \, atm$.
By definition,the compressibility factor is $Z = \frac{PV}{nRT}$.
Therefore,$P = \frac{Z \times nRT}{V} = Z \times P_{ideal}$.
If $Z > 1$,then $P > P_{ideal}$,which means $P > 1 \, atm$.

(C) The Van der Waals equation for a non-ideal gas is given by: $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
In this equation,the term $\frac{an^2}{V^2}$ (or $\frac{a}{V^2}$ for one mole) is added to the pressure $P$ to account for the intermolecular forces of attraction between gas molecules.
The term $b$ represents the excluded volume due to the finite size of gas molecules.

(C) An ideal gas follows the equation $PV = nRT$. We check which of the given options follows this equation. The option that does not follow this equation represents real gas behavior.
$A)$ For $O_2$: $P = 1 \ atm$,$V = 11.2 \ L$,$n = \frac{16}{32} = 0.5 \ mol$,$T = 273 \ K$,$R = 0.0821 \ atm \ L \ K^{-1} \ mol^{-1}$.
$PV = 1 \times 11.2 = 11.2$.
$nRT = 0.5 \times 0.0821 \times 273 \approx 11.2$.
Since $PV = nRT$,it shows ideal behavior.
$B)$ For $H_2$: $P = 24.63 \ atm$,$V = 0.5 \ L$,$n = \frac{1}{2} = 0.5 \ mol$,$T = 300 \ K$.
$PV = 24.63 \times 0.5 = 12.315$.
$nRT = 0.5 \times 0.0821 \times 300 = 12.315$.
Since $PV = nRT$,it shows ideal behavior.
$C)$ For $NH_3$: $P = 1 \ atm$,$V = 22.4 \ L$,$n = 1 \ mol$,$T = 300 \ K$.
$PV = 1 \times 22.4 = 22.4$.
$nRT = 1 \times 0.0821 \times 300 = 24.63$.
Since $PV \neq nRT$,it shows real gas behavior.
$D)$ For $CO_2$: $P = 1 \ atm$,$V = 5.6 \ L$,$n = \frac{11}{44} = 0.25 \ mol$,$T = 273 \ K$.
$PV = 1 \times 5.6 = 5.6$.
$nRT = 0.25 \times 0.0821 \times 273 \approx 5.6$.
Since $PV = nRT$,it shows ideal behavior.

(B) The Van der Waals equation for $1$ mole of gas is $\left( P + \frac{a}{V^2} \right) (V - b) = RT$.
At high pressure,the volume $V$ is small,so the correction term $\frac{a}{V^2}$ becomes very large compared to $P$. However,the term $(V - b)$ cannot be neglected because $V$ is comparable to $b$.
Thus,the pressure correction term $\frac{a}{V^2}$ is neglected relative to $P$,simplifying the equation to $P(V - b) = RT$.

(C) Helium is suitable for gas thermometry at low temperatures because it has a very low boiling point $(4.2 \ K)$ and it behaves almost like an ideal gas even at low temperatures.

(A) Real gases deviate from ideal behavior under conditions of high pressure and low temperature.
This is because at low temperatures,the kinetic energy of molecules is low,making intermolecular forces significant.
At high pressures,the volume occupied by the gas molecules becomes significant relative to the total volume of the container.
Comparing the given options,the condition with the lowest temperature and highest pressure will show the maximum deviation.
Comparing $-10\, ^\circ C$ and $5\, atm$ with the other options,it represents the lowest temperature and the highest pressure combination.
Therefore,the deviation is maximum at $-10\, ^\circ C$ and $5\, atm$.

(B) An ideal gas is defined as a system in which there are no intermolecular or interatomic forces.
Real gases approach ideal gas behaviour at high temperature and low pressure.
Conversely,at low temperature and high pressure,the intermolecular forces become significant and the volume occupied by the gas molecules becomes non-negligible compared to the total volume of the container.
Therefore,the deviation from the ideal gas equation $PV = nRT$ is maximum at low temperature and high pressure.

(A) The compressibility factor is defined as $Z = \frac{P V_m}{R T}$.
Given that $Z > 1$ at $P = 1 \, atm$ and $T = 273 \, K$.
For an ideal gas,$V_{ideal} = \frac{R T}{P} = 22.4 \, L$ at $STP$.
Since $Z = \frac{V_{real}}{V_{ideal}} > 1$,it follows that $V_{real} > V_{ideal}$.
Therefore,$V_m > 22.4 \, L$.

(C) The van der Waals equation for $1 \ mol$ of a gas is $(P + a/V_m^2)(V_m - b) = RT$.
At low pressure,the volume $V_m$ is very large,so the term $b$ (which represents the excluded volume) becomes negligible compared to $V_m$,i.e.,$(V_m - b) \approx V_m$.
Substituting this into the equation,we get $(P + a/V_m^2)(V_m) = RT$.

(A) The compressibility factor is defined as $Z = \frac{V_m}{V_{ideal}}$.
Given that $Z < 1$ at $STP$.
Since the molar volume of an ideal gas at $STP$ is $V_{ideal} = 22.4 \, L$,we have:
$\frac{V_m}{22.4} < 1$
Therefore,$V_m < 22.4 \, L$.

(A) At $STP$,$1 \text{ mole}$ of water vapor occupies $22.4 \text{ L}$ of volume,which includes the space between molecules.
The actual volume occupied by the molecules is the volume of $1 \text{ mole}$ of liquid water $(18 \text{ g})$.
Given the density of water is $1 \text{ g mL}^{-1}$,the volume of $18 \text{ g}$ of water is $18 \text{ mL}$ $(0.018 \text{ L})$.
Comparing this to $22.4 \text{ L}$:
$\text{Percentage} = (0.018 / 22.4) \times 100 \approx 0.08\%$.
Thus,the actual volume is less than $1\%$ of $22.4 \text{ L}$.

(B) The ideal gas equation $PV = nRT$ is followed by real gases under conditions of high temperature and low pressure.
At low temperature,the kinetic energy of gas molecules is low,and intermolecular forces become significant.
At high pressure,the volume occupied by the gas molecules becomes significant compared to the total volume of the container.
Therefore,real gases deviate most from ideal behavior at low temperature and high pressure.

(C) real gas behaves most like an ideal gas under conditions of low pressure and high temperature.
At low pressure,the volume occupied by the gas molecules becomes negligible compared to the total volume of the container.
At high temperature,the kinetic energy of the gas molecules is high,making the intermolecular forces of attraction negligible.
Comparing the given options,$0.5 \, atm$ (lowest pressure) and $500 \, K$ (highest temperature) provide the conditions closest to ideal behavior.

(B) The van der Waals equation is given by $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
This equation was derived to account for the deviations of real gases from ideal behavior.
It incorporates corrections for the finite volume of gas molecules $(b)$ and the intermolecular forces of attraction $(a)$.
Therefore,it is specifically valid for real gases.

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

(A) The pressure exerted by a gas on the walls of a container is defined as the force per unit area. According to the kinetic theory of gases and the $van \ der \ Waals$ model,this pressure arises from the continuous bombardment of gas molecules against the container walls. Specifically,the pressure depends on two main factors:
$1$. The frequency of collisions of the molecules with the wall.
$2$. The change in momentum of the molecules upon collision with the wall,which exerts a force according to Newton's second law $(F = \frac{dp}{dt})$.
Therefore,the pressure is a direct result of the collision frequency and the momentum transfer.

(B) The temperature at which a real gas behaves like an ideal gas over an appreciable range of pressure is known as the $Boyle$ temperature or $Boyle$ point $(T_B)$.
At this temperature,the compressibility factor $Z$ remains close to $1$ for a wide range of pressure.

(B) According to the kinetic molecular theory of gases,an ideal gas assumes that there are no intermolecular forces of attraction between gas molecules.
However,in real gases,there are weak intermolecular forces of attraction (van der Waals forces) between the molecules.
These forces cause real gases to deviate from ideal behavior,especially at high pressures and low temperatures.

(B) The van der Waals equation is given by $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
For $1 \text{ mole}$ of gas,it becomes $(P + \frac{a}{V^2})(V - b) = RT$.
At low pressure,the volume $V$ is very large,so $b$ is negligible compared to $V$.
At high temperature,the term $\frac{a}{V^2}$ becomes negligible because $V$ is large.
Thus,the equation simplifies to $PV = RT$,which is the ideal gas equation.

(B) The van der Waals equation for a real gas is: $(P + \frac{an^2}{V^2})(V - nb) = nRT$
Given: $n = 5 \ mol$,$V = 1 \ L$,$T = 47 + 273 = 320 \ K$,$a = 3.592 \ atm \ L^2 \ mol^{-2}$,$b = 0.0427 \ L \ mol^{-1}$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$
Substituting the values: $(P + \frac{3.592 \times 5^2}{1^2})(1 - 5 \times 0.0427) = 5 \times 0.0821 \times 320$
$(P + 89.8)(1 - 0.2135) = 131.36$
$(P + 89.8)(0.7865) = 131.36$
$P + 89.8 = \frac{131.36}{0.7865} \approx 167.02$
$P = 167.02 - 89.8 = 77.22 \ atm$
Thus,the pressure is approximately $77.2 \ atm$.

(B) The compressibility factor $Z$ is given by the formula $Z = \frac{PV}{nRT}$.
Given: $P = 40 \ atm$,$V = 0.4 \ L$,$n = 1 \ mol$,$R = 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,$T = 300 \ K$.
Substituting the values: $Z = \frac{40 \times 0.4}{1 \times 0.0821 \times 300} = \frac{16}{24.63} \approx 0.65$.
Since $Z < 1$,the attractive forces are dominant,which means the gas is more compressible than an ideal gas.

(C) The van der Waals constant $'a'$ represents the magnitude of attractive forces between gas molecules.
Greater the intermolecular forces,higher is the value of $'a'$.
Among the given gases,$NH_3$ is a polar molecule capable of forming hydrogen bonds,which results in strong intermolecular attractive forces.
Therefore,$NH_3$ has the highest value of $'a'$ compared to the non-polar gases $He$,$H_2$,and $O_2$.

(A) The van der Waals equation for $1 \text{ mole}$ of a real gas is given by: $(P + \frac{a}{V_m^2})(V_m - b) = RT$.
In this equation,the term $\frac{a}{V_m^2}$ accounts for the intermolecular attractive forces between gas molecules.
The term $(V_m - b)$ accounts for the excluded volume due to the finite size of the gas molecules.
Therefore,the term representing intermolecular forces is $P + \frac{a}{V_m^2}$.

(C) The compressibility factor $Z$ is defined as $Z = \frac{PV_m}{RT}$.
For an ideal gas,$Z = 1$ and the molar volume $V_m$ at $STP$ is $22.4 \ dm^3$.
Given that $Z > 1$,the gas shows positive deviation from ideal behavior,meaning the repulsive forces are dominant.
Since $Z = \frac{V_{m, real}}{V_{m, ideal}}$,if $Z > 1$,then $V_{m, real} > V_{m, ideal}$.
Therefore,the molar volume $V_m$ will be greater than $22.4 \ dm^3$.

(C) The van der Waals equation for $1 \ mol$ of a gas is given by: $\left( P + \frac{a}{V_m^2} \right) (V_m - b) = RT$.
At low pressure,the volume $V_m$ is very large,so $V_m \gg b$. Therefore,the term $(V_m - b)$ can be approximated as $V_m$.
Substituting this into the equation: $\left( P + \frac{a}{V_m^2} \right) V_m = RT$.
Expanding this,we get $PV_m + \frac{a}{V_m} = RT$.
Rearranging for $P$,we get $P = \frac{RT}{V_m} - \frac{a}{V_m^2}$.
However,the standard simplified form at low pressure is often represented as $\left( P + \frac{a}{V_m^2} \right) V_m = RT$.

(A) The van der Waals' equation for one mole of gas is $(P + \frac{a}{V^2})(V - b) = RT$.
At low pressures,the volume $V$ is large,so $b$ is negligible compared to $V$.
The equation simplifies to $(P + \frac{a}{V^2})V = RT$.
Expanding this,we get $PV + \frac{a}{V} = RT$.
Dividing both sides by $RT$,we get $\frac{PV}{RT} + \frac{a}{VRT} = 1$.
Since the compressibility factor $Z = \frac{PV}{RT}$,we have $Z + \frac{a}{VRT} = 1$.
Therefore,$Z = 1 - \frac{a}{VRT}$.

(B) The general Vander Waals equation for $n$ moles of a gas is given by: $\left( P + \frac{an^2}{V^2} \right) \left( V - nb \right) = nRT$
Substituting $n = 0.5 = \frac{1}{2}$ mol into the equation:
$\left( P + \frac{a(1/2)^2}{V^2} \right) \left( V - \frac{b}{2} \right) = \frac{1}{2} RT$
$\left( P + \frac{a}{4V^2} \right) \left( V - \frac{b}{2} \right) = \frac{RT}{2}$
Multiplying both sides by $2$:
$\left( P + \frac{a}{4V^2} \right) \left( 2V - b \right) = RT$

(B) Boyle's temperature is the temperature at which a real gas obeys the ideal gas equation over an appreciable range of pressure.
At this temperature,the compressibility factor $Z$ is equal to $1$.

(A) The given equation is $\left[ P + \frac{a}{V^2} \right] V = RT$.
Expanding the equation: $PV + \frac{a}{V} = RT$.
Dividing both sides by $RT$: $\frac{PV}{RT} + \frac{a}{RTV} = 1$.
Since the compressibility factor is defined as $Z = \frac{PV}{RT}$,we substitute this into the equation: $Z + \frac{a}{RTV} = 1$.
Rearranging for $Z$,we get: $Z = 1 - \frac{a}{RTV}$.

(A) The Van der Waals equation for $n$ moles of a real gas is given by: $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
For a real gas to behave like an ideal gas,the deviations from ideal behavior must be minimized.
$1$. The term $\frac{an^2}{V^2}$ accounts for intermolecular forces of attraction. If $a$ is small,these forces are negligible.
$2$. The term $nb$ accounts for the finite volume occupied by gas molecules. If $b$ is small,the volume of the molecules is negligible compared to the total volume of the container.
Therefore,when both Van der Waals constants $a$ and $b$ are small,the real gas equation reduces to the ideal gas equation $PV = nRT$.

(C) real gas behaves as an ideal gas under conditions of low pressure and high temperature.
Conversely,a real gas deviates from ideal behaviour at high pressure and low temperature.
Therefore,the correct conditions for deviation are $B$ (High pressure) and $C$ (Low temperature).