Ideal gas equation and Related gas laws Questions in English

(A) Normal temperature and pressure $(NTP)$ implies $P_1 = 1 \ atm = 1.01325 \times 10^5 \ Nm^{-2}$ and $V_1 = 1 \ dm^3$.
Given $P_2 = 1.032 \times 10^5 \ Nm^{-2}$.
Using Boyle's Law,$P_1 V_1 = P_2 V_2$.
$V_2 = \frac{P_1 V_1}{P_2} = \frac{1.01325 \times 10^5 \ Nm^{-2} \times 1 \ dm^3}{1.032 \times 10^5 \ Nm^{-2}}$.
$V_2 \approx 0.982 \ dm^3$.

(A) Given: $n = 3.4 \ mol$,$T = 300 \ K$,$V = 68 \ mL = 68 \times 10^{-3} \ L = 0.068 \ dm^3$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Using the ideal gas equation: $PV = nRT$.
$P = \frac{nRT}{V} = \frac{3.4 \ mol \times 8.314 \ J \ K^{-1} \ mol^{-1} \times 300 \ K}{0.068 \ dm^3}$.
Since $1 \ J = 1 \ Pa \cdot m^3$ and $1 \ dm^3 = 10^{-3} \ m^3$,the pressure $P = \frac{8480.28}{0.068 \times 10^{-3}} \ Pa = 124710000 \ Pa = 1.2471 \times 10^5 \ kPa$ is incorrect in the original prompt's unit conversion.
Recalculating: $P = \frac{3.4 \times 8.314 \times 300}{0.068 \times 10^{-3}} \ Pa = 1.2471 \times 10^8 \ Pa = 1.2471 \times 10^5 \ kPa$.

(C) Boyle's law states that at constant temperature,$PV = k_1$.
Charles's law states that at constant pressure,$\frac{V}{T} = k_2$.
Combining these two laws for a fixed amount of gas,we get the combined gas law: $\frac{PV}{T} = k$.
For a gas changing from state $1$ $(P_1, V_1, T_1)$ to state $2$ $(P_2, V_2, T_2)$,the relationship is expressed as:
$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.

(B) The ideal gas constant $R$ is a fundamental physical constant.
In the units of $L \ atm \ K^{-1} \ mol^{-1}$,the value of $R$ is approximately $0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
Among the given options,$0.082$ is the closest approximation to this value.
Therefore,the correct option is $B$.

(C) According to Charles' law,$\frac{V_1}{T_1} = \frac{V_2}{T_2}$ (at constant $n$ and $P$).
Given: $V_1 = 22.4 \ L$,$T_1 = 0^{\circ} C = 273 \ K$,$V_2 = 224 \ L$.
Substituting the values into the formula:
$T_2 = \frac{V_2 \times T_1}{V_1} = \frac{224 \times 273}{22.4} = 10 \times 273 = 2730 \ K$.

(B) According to Charles' law,at constant pressure,the volume of a fixed mass of gas is directly proportional to its absolute temperature: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = 3 \ dm^3$,$T_1 = 300 \ K$,and $V_2 = 2 \times V_1 = 6 \ dm^3$.
Substituting the values into the equation: $\frac{3 \ dm^3}{300 \ K} = \frac{6 \ dm^3}{T_2}$.
Solving for $T_2$: $T_2 = \frac{6 \ dm^3 \times 300 \ K}{3 \ dm^3} = 600 \ K$.

(C) According to Boyle's law, for a fixed amount of gas at constant temperature, $P_1 V_1 = P_2 V_2$.
Given: $P_1 = 2 \text{ atm}$, $V_1 = 9.0 \text{ L}$, $V_2 = 3.0 \text{ L}$.
Substituting the values: $2 \text{ atm} \times 9.0 \text{ L} = P_2 \times 3.0 \text{ L}$.
$P_2 = \frac{2 \times 9.0}{3.0} \text{ atm} = 6.0 \text{ atm}$.

(D) According to Charles's Law, at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature $(V \propto T)$.
Given:
Initial volume $(V_1)$ = $2000 \text{ dm}^3$
Initial temperature $(T_1)$ = $99^\circ\text{C} = 99 + 273 = 372 \text{ K}$
Final temperature $(T_2)$ = $80^\circ\text{C} = 80 + 273 = 353 \text{ K}$
Using the formula: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$
$V_2 = \frac{V_1 \times T_2}{T_1}$
$V_2 = \frac{2000 \times 353}{372}$
$V_2 \approx 1897.8 \text{ dm}^3$.

(C) According to Charles' law,at constant pressure and amount of gas,the volume is directly proportional to the absolute temperature: $V_1 / T_1 = V_2 / T_2$.
Initial temperature $T_1 = 0^{\circ} C = 273 \ K$.
Final temperature $T_2 = 0^{\circ} C + 10^{\circ} C = 10^{\circ} C = 283 \ K$.
Given $V_1 = 4 \ dm^3$.
Substituting the values: $4 \ dm^3 / 273 \ K = V_2 / 283 \ K$.
$V_2 = (4 \times 283) / 273 \approx 4.1465 \ dm^3$.
Rounding to two decimal places,the volume is $4.14 \ dm^3$.

(B) According to the ideal gas equation,$PV = nRT$.
Substituting the given values: $P = 4.926 \ atm$,$V = 20 \ dm^3$,$n = 2 \ mol$,and $R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$.
$T = \frac{PV}{nR} = \frac{4.926 \times 20}{2 \times 0.0821} = \frac{98.52}{0.1642} = 600 \ K$.
To convert the temperature to degree Celsius: $T(^{\circ}C) = T(K) - 273 = 600 - 273 = 327^{\circ}C$.

(D) According to Boyle's Law,at constant temperature,$P_1 V_1 = P_2 V_2$.
Given: $P_1 = 1 \ atm$,$V_1 = 60 \ mL$,$P_2 = 1.5 \ atm$.
Substituting the values: $1 \ atm \times 60 \ mL = 1.5 \ atm \times V_2$.
$V_2 = \frac{60}{1.5} \ mL = 40 \ mL$.
Since $1 \ dm^3 = 1000 \ mL$,$40 \ mL = 40 \times 10^{-3} \ dm^3 = 4 \times 10^{-2} \ dm^3$.

(A) Using the ideal gas equation: $PV = nRT$
Given:
$n = 1.5 \ mol$
$V = 3 \ dm^3$
$T = 300 \ K$
$R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$
Substituting the values:
$P \times 3 = 1.5 \times 0.0821 \times 300$
$P \times 3 = 36.945$
$P = \frac{36.945}{3} = 12.315 \ atm \approx 12.32 \ atm$.

(B) Using the ideal gas equation: $PV = nRT$
Given: $P = 4.5 \ atm$,$V = 2.5 \ L$,$T = 300 \ K$,and $R = 0.0821 \ atm \ L \ K^{-1} \ mol^{-1}$.
Substituting the values: $4.5 \times 2.5 = n \times 0.0821 \times 300$
$11.25 = n \times 24.63$
$n = \frac{11.25}{24.63} \approx 0.4567 \ mol$
Rounding to two decimal places,we get $n \approx 0.46 \ mol$.

(A) Gay-Lussac's law states that the pressure of a given mass of gas is directly proportional to its absolute temperature,provided the volume remains constant.
Mathematically,this is expressed as $P \propto T$ at constant volume and mass.

(D) According to Boyle's Law,at constant temperature,$P_1 V_1 = P_2 V_2$.
Given: $P_1 = 105 \ kPa$,$V_1 = 11.2 \ dm^3$,$P_2 = 420 \ kPa$.
Substituting the values: $105 \times 11.2 = 420 \times V_2$.
$V_2 = \frac{105 \times 11.2}{420}$.
$V_2 = 2.8 \ dm^3$.

(C) Given: $P_1 = 105 \ kPa$,$V_1 = 11.2 \ dm^3$,$P_2 = 210 \ kPa$.
According to Boyle's law,at constant temperature,$P_1 \ V_1 = P_2 \ V_2$.
Substituting the values: $105 \ kPa \times 11.2 \ dm^3 = 210 \ kPa \times V_2$.
$V_2 = \frac{105 \ kPa \times 11.2 \ dm^3}{210 \ kPa} = 5.6 \ dm^3$.

(D) According to Charles's Law,$\frac{V_1}{T_1} = \frac{V_2}{T_2}$ at constant pressure.
Given: $V_1 = 2 \ dm^3$,$T_1 = 0^{\circ} C = 273 \ K$.
The temperature is decreased by $272^{\circ} C$,so the new temperature $T_2 = 273 \ K - 272 \ K = 1 \ K$.
Substituting the values: $\frac{2}{273} = \frac{V_2}{1}$.
Therefore,$V_2 = \frac{2}{273} \ dm^3$.

(C) Given: $P = 5 \ atm$,$T = 300 \ K$,$mass = 22 \ g$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
Number of moles $(n)$ of $CO_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{22 \ g}{44 \ g \ mol^{-1}} = 0.5 \ mol$.
Using the ideal gas equation: $PV = nRT$.
$V = \frac{nRT}{P} = \frac{0.5 \times 0.0821 \times 300}{5}$.
$V = \frac{12.315}{5} = 2.463 \ L$.
Since $1 \ L = 1 \ dm^3$,the volume is $2.46 \ dm^3$.

(B) According to Boyle's Law,at constant temperature,$P_1 V_1 = P_2 V_2$.
Given: $P_1 = 1.013 \times 10^5 \ Nm^{-2}$,$V_1 = 2.27 \ L$.
Final volume $V_2 = 4540 \ mL = 4.540 \ L$.
Substituting the values: $1.013 \times 10^5 \times 2.27 = P_2 \times 4.540$.
$P_2 = \frac{1.013 \times 10^5 \times 2.27}{4.540} = 0.5065 \times 10^5 \ Nm^{-2} = 5.065 \times 10^4 \ Nm^{-2}$.

(A) According to Charles's Law,for a fixed amount of gas at constant pressure,the volume is directly proportional to the temperature: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = 3.4 \ L$,$T_1 = 298 \ K$,$V_2 = 6.8 \ L$.
Substituting the values: $\frac{3.4}{298} = \frac{6.8}{T_2}$.
Solving for $T_2$: $T_2 = \frac{6.8 \times 298}{3.4} = 2 \times 298 = 596 \ K$.

(C) According to Boyle's Law,at constant temperature,$P_1 V_1 = P_2 V_2$.
Given: $P_1 = 2 \ atm$,$V_1 = 250 \ mL$,$P_2 = 2.5 \ atm$.
Substituting the values: $2 \times 250 = 2.5 \times V_2$.
$V_2 = \frac{2 \times 250}{2.5} = \frac{500}{2.5} = 200 \ mL$.

(A) Given: $n = 2 \ mol$,$T = 546 \ K$,$V = 44.8 \ L$,$R = 0.0821 \ L \ atm \ mol^{-1} \ K^{-1}$.
According to the ideal gas equation,$PV = nRT$.
Therefore,$P = \frac{nRT}{V}$.
Substituting the values: $P = \frac{2 \times 0.0821 \times 546}{44.8}$.
$P = \frac{89.6532}{44.8} \approx 2.0 \ atm$.

(C) Given: $V_{1} = 0.2 \ L$,$T_{1} = 273.15 \ K$.
The final temperature is $T_{2} = 273.15 \ ^{\circ}C = 273.15 + 273.15 = 546.30 \ K$.
According to Charles's Law,at constant pressure $(P)$ and amount of gas $(n)$:
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
Substituting the values:
$V_{2} = \frac{V_{1} \times T_{2}}{T_{1}} = \frac{0.2 \times 546.30}{273.15} = 0.4 \ L$.

(C) Given: $V_{1} = 3.4 \ L$,$T_{1} = 25^{\circ} C = 25 + 273 = 298 \ K$.
$V_{2} = 10.2 \ L$,$T_{2} = ?$.
According to Charles's Law,at constant pressure and amount of gas,$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$.
Substituting the values: $\frac{3.4 \ L}{298 \ K} = \frac{10.2 \ L}{T_{2}}$.
$T_{2} = \frac{10.2 \ L \times 298 \ K}{3.4 \ L} = 3 \times 298 \ K = 894 \ K$.

(D) Given: $V_{1} = 2 \ dm^{3}$,$T_{1} = 273.15 \ K$ (at $STP$).
We need to find $T_{2}$ when $V_{2} = 2 \times V_{1} = 4 \ dm^{3}$ at constant pressure.
According to Charles's law,$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$.
Substituting the values: $\frac{2}{273.15} = \frac{4}{T_{2}}$.
$T_{2} = \frac{4 \times 273.15}{2} = 546.3 \ K$.
To convert to Celsius: $t^{\circ} C = T(K) - 273.15$.
$t^{\circ} C = 546.3 - 273.15 = 273.15^{\circ} C$.

(A) According to Boyle's law,at constant temperature,$P_1 V_1 = P_2 V_2$.
Given: $P_1 = 720 \ mm$,$V_1 = 100 \ mL$,$V_2 = 84 \ mL$.
Substituting the values into the equation: $P_2 = \frac{P_1 V_1}{V_2}$.
$P_2 = \frac{720 \times 100}{84} = \frac{72000}{84} \approx 857.14 \ mm$.

(A) Given: $n=2 \ mol$,$T=546 \ K$,$V=44.8 \ L$,$R=0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.
According to the ideal gas equation,$PV=nRT$.
Substituting the values: $P = \frac{nRT}{V} = \frac{2 \times 0.082 \times 546}{44.8}$.
$P = \frac{89.544}{44.8} \approx 1.998 \ atm$.
Therefore,the correct option is $A$.

(D) According to Charles's Law,the volume of a fixed mass of gas is directly proportional to its absolute temperature $(V \propto T)$.
As the temperature decreases,the volume of the gas decreases.
At a temperature of $-273.15^{\circ} C$ (which is $0 \ K$),the volume of a gas theoretically becomes zero.
This temperature is known as absolute zero.

(B) Given: $P_{1} = 1 \ bar$,$V_{1} = 2.27 \ L$,$P_{2} = 0.227 \ bar$.
According to Boyle's Law,$P_{1}V_{1} = P_{2}V_{2}$ at constant temperature.
Substituting the values: $1 \ bar \times 2.27 \ L = 0.227 \ bar \times V_{2}$.
$V_{2} = \frac{1 \times 2.27}{0.227} = 10 \ L$.
The rise in volume of the gas is $\Delta V = V_{2} - V_{1} = 10 \ L - 2.27 \ L = 7.73 \ L$.

(A) Let the mass of each gas be $m = 1 \ g$.
The number of moles $n = \frac{m}{M}$,where $M$ is the molar mass.
Since $m$ is constant,$n \propto \frac{1}{M}$.
The molar masses are: $M(H_2) = 2 \ g/mol$,$M(N_2) = 28 \ g/mol$,$M(O_2) = 32 \ g/mol$,$M(Cl_2) = 71 \ g/mol$.
Thus,the number of moles follows the order: $n(H_2) > n(N_2) > n(O_2) > n(Cl_2)$.
From the ideal gas equation $PV = nRT$,at constant $T$ and $V$,$P \propto n$.
Therefore,the gas with the highest number of moles will exert the highest pressure.
Since $H_2$ has the lowest molar mass,it has the highest number of moles for a given mass,and thus exerts the highest pressure.

(A) The work done during the compression of a gas at constant pressure is given by the formula: $W = -P_{ext} \Delta V$.
Given: $P_{ext} = 0.2 \ bar$,$V_1 = 1 \ m^3$,$V_2 = 0.5 \ m^3$.
$\Delta V = V_2 - V_1 = 0.5 \ m^3 - 1 \ m^3 = -0.5 \ m^3$.
$W = -(0.2 \ bar) \times (-0.5 \ m^3) = 0.1 \ bar \cdot m^3$.
Since $1 \ bar = 10^5 \ Pa$ and $1 \ Pa \cdot m^3 = 1 \ J$,then $1 \ bar \cdot m^3 = 10^5 \ J = 100 \ kJ$.
Therefore,$W = 0.1 \times 100 \ kJ = 10 \ kJ$.

(A) According to Charles's Law,at constant pressure,the volume of a fixed mass of gas is directly proportional to its absolute temperature: $V_1 / T_1 = V_2 / T_2$.
Given: $V_1 = 4 \ dm^3$,$T_1 = 273 \ K$,$T_2 = 546 \ K$.
Substituting the values: $4 / 273 = V_2 / 546$.
$V_2 = (4 \times 546) / 273 = 4 \times 2 = 8 \ dm^3$.

(C) Given: $V_1 = 10 \ L$,$V_2 = 5 \ L$,$W = 1730 \ J$,$T = 300 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
For isothermal reversible compression,the work done on the system is given by the formula:
$W = -2.303 \ nRT \log_{10} \left(\frac{V_2}{V_1}\right)$
Since work is done on the gas,$W$ is positive in this context $(W = 1730 \ J)$:
$1730 = -2.303 \times n \times 8.314 \times 300 \times \log_{10} \left(\frac{5}{10}\right)$
$1730 = -2.303 \times n \times 8.314 \times 300 \times \log_{10} (0.5)$
$1730 = -2.303 \times n \times 8.314 \times 300 \times (-0.3010)$
$1730 = n \times 5744.14 \times 0.3010$
$1730 = n \times 1729$
$n = \frac{1730}{1729} \approx 1 \ mol$.

(A) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
For a monoatomic ideal gas,the degrees of freedom $f = 3$.
The adiabatic index $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$.
Thus,$\gamma - 1 = \frac{2}{3}$.
Since the volume $V$ is proportional to the length $L$ of the gas column ($V = A \times L$,where $A$ is the constant cross-sectional area),we have $\frac{V_1}{V_2} = \frac{L_1}{L_2}$.
Substituting this into the adiabatic relation: $\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1} = \left(\frac{L_1}{L_2}\right)^{2/3}$.

(B) For an isothermal reversible expansion of an ideal gas,the work done by the gas is given by:
$W = nRT \ln\left(\frac{V_2}{V_1}\right)$
Given:
$W = 5.187 \ kJ = 5187 \ J$
$V_1 = 10 \ m^3$
$V_2 = 20 \ m^3$
$T = 300 \ K$
$R = 8.314 \ J \ mol^{-1} \ K^{-1}$
Substituting the values:
$5187 = n \times 8.314 \times 300 \times \ln\left(\frac{20}{10}\right)$
$5187 = n \times 8.314 \times 300 \times 0.6931$
$5187 = n \times 1728.5$
$n = \frac{5187}{1728.5} \approx 3$
Therefore,the number of moles of gas used is $3$.

(C) Given: $V_{1} = 10 \ m^{3}$,$V_{2} = 20 \ m^{3}$,$T = 300 \ K$,$W = -5.187 \ kJ = -5187 \ J$.
For an isothermal reversible expansion,the work done is given by the formula:
$W = -2.303 \ nRT \log_{10} \left(\frac{V_{2}}{V_{1}}\right)$
Substituting the values:
$-5187 = -2.303 \times n \times 8.314 \times 300 \times \log_{10} \left(\frac{20}{10}\right)$
$-5187 = -2.303 \times n \times 8.314 \times 300 \times 0.3010$
$n = \frac{5187}{2.303 \times 8.314 \times 300 \times 0.3010}$
$n = \frac{5187}{1729.0} \approx 3 \ mol$.

(A) For an ideal gas,the work done in an expansion process is related to the number of moles $n$ by the ideal gas equation $pV = nRT$.
Since $p$,$V$,and $T$ are constant,$n$ must be constant for all gases.
Given that the mass $m$ of each gas is the same,the number of moles is given by $n = \frac{m}{M}$,where $M$ is the molar mass.
For $n$ to be constant for a fixed mass $m$,the molar mass $M$ must be the same,which is not possible for different gases.
However,if we consider the work done per unit mass or the expansion work capacity,$W \propto n = \frac{m}{M}$.
Since $m$ is constant,$W \propto \frac{1}{M}$.
Comparing the molar masses: $M(NH_{3}) = 17 \ g/mol$,$M(N_{2}) = 28 \ g/mol$,$M(Cl_{2}) = 71 \ g/mol$,and $M(H_{2}S) = 34 \ g/mol$.
Since $NH_{3}$ has the lowest molar mass,it will have the highest number of moles for a given mass,resulting in the maximum work done.

(B) From the ideal gas equation,density $\rho = \frac{PM}{RT}$.
Since $M$ (molar mass) and $R$ (gas constant) are constant for carbon monoxide,we have $\rho \propto \frac{P}{T}$.
We calculate the ratio $\frac{P}{T}$ for each option:

$A$. $0.5 \ atm, 273 \ K$	$\frac{0.5}{273} \approx 0.0018$
$B$. $4 \ atm, 500 \ K$	$\frac{4}{500} = 0.008$
$C$. $2 \ atm, 600 \ K$	$\frac{2}{600} \approx 0.0033$
$D$. $6 \ atm, 1092 \ K$	$\frac{6}{1092} \approx 0.0055$

The ratio $\frac{P}{T}$ is highest for option $B$. Thus,the density is maximum at $4 \ atm$ and $500 \ K$.

(C) The gram molecular volume is defined as the volume occupied by $1 \ \text{mole}$ of any gas at $STP$ $(Standard \ Temperature \ and \ Pressure)$.
According to Avogadro's hypothesis,$1 \ \text{mole}$ of any ideal gas occupies $22.4 \ L$ or $22400 \ cm^{3}$ at $STP$.
Therefore,the gram molecular volume of oxygen at $STP$ is $22400 \ cm^{3}$.

(C) According to Boyle's law,at constant temperature,$P_1 V_1 = P_2 V_2$.
Let the initial volume be $V_1 = V$ and initial pressure be $P_1 = P$.
To increase the volume by $10 \%$,the new volume $V_2 = V + 0.1V = 1.1V$.
Substituting these values in the equation: $P \times V = P_2 \times 1.1V$.
$P_2 = \frac{P}{1.1} \approx 0.909P$.
The change in pressure is $P_2 - P_1 = 0.909P - P = -0.091P$.
This represents a decrease of approximately $9.1 \%$.
However,in the context of simple proportional relationships often tested in this format,if $V$ increases by $10 \%$,the pressure must decrease. Among the given options,the most appropriate choice based on the inverse relationship $P \propto \frac{1}{V}$ is a decrease.

(B) Given: $p_1 = 2 \ atm$,$T_1 = 25 + 273 = 298 \ K$,$T_2 = 323 + 273 = 596 \ K$.
Let initial volume $V_1 = V$. Then final volume $V_2 = \frac{2}{3} V$.
Using the combined gas law: $\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$.
Substituting the values: $\frac{2 \times V}{298} = \frac{p_2 \times (2/3) V}{596}$.
$p_2 = \frac{2 \times 596}{298 \times (2/3)} = \frac{2 \times 2}{2/3} = 2 \times 3 = 6 \ atm$.

(B) Given,mass of $CO = 2.8 \ g$.
Temperature $T = 27 + 273 = 300 \ K$.
Pressure $p = 0.821 \ atm$.
Gas constant $R = 0.08210 \ L \ atm \ K^{-1} \ mol^{-1}$.
Molar mass of $CO = 12 + 16 = 28 \ g \ mol^{-1}$.
Number of moles $n = \frac{2.8 \ g}{28 \ g \ mol^{-1}} = 0.1 \ mol$.
Using the ideal gas equation $pV = nRT$:
$V = \frac{nRT}{p} = \frac{0.1 \times 0.08210 \times 300}{0.821} = \frac{2.463}{0.821} = 3 \ L$.

(B) According to the ideal gas law,$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$.
Given: $T_{2} = 2 T_{1}$ and $p_{2} = \frac{p_{1}}{2}$.
Substituting these values into the equation:
$\frac{p_{1} V_{1}}{T_{1}} = \frac{(p_{1} / 2) \times V_{2}}{2 T_{1}}$.
Canceling $p_{1}$ and $T_{1}$ from both sides:
$V_{1} = \frac{V_{2}}{4}$.
Therefore,$V_{2} = 4 V_{1}$.
The volume becomes $4$ times the original volume.

(A) Charles's law states that for a given mass of gas at constant pressure,the volume $V$ is directly proportional to the absolute temperature $T$,expressed as $V/T = K$ (where $K$ is a constant).
Taking the natural logarithm on both sides: $\ln V = \ln K + \ln T$.
Taking the common logarithm on both sides: $\log V = \log K + \log T$.
Rearranging the logarithmic equation: $\log V - \log T = \log K$.
Option $A$ represents $\log K = \log V + \log T$,which implies $\log K = \log(VT)$,or $K = VT$. This contradicts Charles's law $(V/T = K)$.
Therefore,option $A$ does not represent Charles's law.

(C) According to the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$.
Since $P$,$V$,$T$,and $m$ are given,$V = \frac{mRT}{PM}$.
For a fixed mass $(m = 25 \ g)$,temperature $(T)$,and pressure $(P)$,the volume $V$ is inversely proportional to the molar mass $M$ $(V \propto \frac{1}{M})$.
The molar masses of the gases are: $M(HF) = 20 \ g/mol$,$M(HCl) = 36.5 \ g/mol$,$M(HBr) = 81 \ g/mol$,and $M(HI) = 128 \ g/mol$.
Since $HI$ has the highest molar mass,it will have the least number of moles and consequently the least volume.

(A) For an ideal gas,the equation of state is $PV = nRT$.
The compressibility factor $(Z)$ is defined as $Z = \frac{PV}{nRT}$.
Since $PV = nRT$ for an ideal gas,the value of $Z$ is always $1$.

(C) Given: Volume $V = 100 \ L$,Final pressure $P_2 = 3 \ atm$,Temperature $T = 300 \ K$.
Using the ideal gas law $PV = nRT$:
$n_2 = \frac{P_2 V}{RT}$
Substituting the values:
$n_2 = \frac{3 \ atm \times 100 \ L}{R \times 300 \ K} = \frac{300}{300 R} = \frac{1}{R} \ mol$.
Thus,the number of moles of $H_2$ remaining in the cylinder is $\frac{1}{R}$.

(A) Using the ideal gas equation: $PV = nRT = \frac{m}{M}RT$
Rearranging for molar mass $M$: $M = \frac{mRT}{PV}$
Substituting the given values: $M = \frac{(1 \ g)(0.082 \ L \ atm \ mol^{-1} \ K^{-1})(1000 \ K)}{(0.5 \ atm)(2.5625 \ L)}$
$M = \frac{82}{1.28125} = 64 \ g \ mol^{-1}$.

(C) The ideal gas equation is given by $PV = nRT$.
For option $A$: If $n$ and $T$ are constant, then $PV = \text{constant}$, which means $P \propto 1/V$. The graph of $P$ versus $V$ should be a rectangular hyperbola, not a straight line.
For option $B$: If $V$ and $n$ are constant, then $P = (nR/V) \times T$. Since $nR/V$ is constant, $P \propto T$. The graph of $P$ versus $T$ should be a straight line passing through the origin, not a line with a negative slope.
For option $C$: If $V$ and $T$ are constant, then $P = (RT/V) \times n$. Since $RT/V$ is constant, $P \propto n$. The graph of $P$ versus $n$ is a straight line passing through the origin. This is correct.
For option $D$: If $P$ and $n$ are constant, then $V = (nR/P) \times T$. Thus $V \propto T$. The graph of $V$ versus $1/T$ would not be a straight line passing through the origin; rather, $V$ versus $T$ would be.
Therefore, the correct graph is $C$.

(C) For an ideal gas,the equation is $PV = nRT$.
For an isochore (constant volume process),the equation can be written as $P = (\frac{nR}{V})T$.
The slope of the $P$ vs $T$ graph is given by $m = \frac{nR}{V}$.
Given $n = 1 \ mol$ and slope $m = 0.082 \ atm \ K^{-1}$.
So,$\frac{1 \times 0.082}{V} = 0.082$,which implies $V = 1 \ L$.
Now,at $T = 12.2 \ K$,the pressure $P$ is calculated as $P = \frac{nRT}{V} = \frac{1 \times 0.082 \times 12.2}{1} = 1.0004 \ atm \approx 1.0 \ atm$.