Ideal gas equation and Related gas laws Questions in English

(B) According to Boyle's Law,at constant temperature,$P_1 V_1 = P_2 V_2$.
Given: $P_1 = 9.962 \times 10^4 \, N m^{-2}$,$V_1 = 95 \, cm^3$,$P_2 = 10.13 \times 10^4 \, N m^{-2}$.
Substituting the values: $V_2 = \frac{P_1 V_1}{P_2} = \frac{9.962 \times 10^4 \times 95}{10.13 \times 10^4}$.
$V_2 = 93.4 \, cm^3$.

(A) The ideal gas equation is $PV = nRT$,which gives $R = \frac{PV}{nT}$.
At $STP$,for $1 \, mol$ of an ideal gas,the pressure $P = 1 \, atm$,volume $V = 22.4 \, L$,and temperature $T = 273.15 \, K$.
Substituting these values: $R = \frac{1 \, atm \times 22.4 \, L}{1 \, mol \times 273.15 \, K} \approx 0.082 \, L \, atm \, K^{-1} \, mol^{-1}$.

(C) We know the ideal gas equation is $PM = dRT$,where $P$ is pressure,$M$ is molar mass,$d$ is density,$R$ is the gas constant,and $T$ is temperature.
At constant pressure and for a given gas,$d \times T = \text{constant}$.
Therefore,$d_1 T_1 = d_2 T_2$.
Given: $d_1 = d$,$T_1 = 27\,^oC = (273 + 27)\,K = 300\,K$,and $d_2 = 0.75\,d$.
Substituting the values: $d \times 300 = 0.75\,d \times T_2$.
$T_2 = \frac{300}{0.75} = 400\,K$.

(C) From the ideal gas equation,we know that $PM = dRT$.
Here,$P = 2.0\,atm$,$M_{CH_4} = 16\,g\,mol^{-1}$,$R = 0.0821\,L\,atm\,K^{-1}mol^{-1}$,and $T = 27 + 273 = 300\,K$.
Substituting these values:
$d = \frac{PM}{RT} = \frac{2.0 \times 16}{0.0821 \times 300} \approx 1.30\,g\,L^{-1}$.

(B) From the ideal gas equation $PV = nRT$,it can be concluded that $n$ moles of gas at temperature $T$ and pressure $P$ occupy a volume $V$.

(C) At $STP$,the molar volume of an ideal gas is $22.4 \ L \ mol^{-1}$.
The number of moles $n$ is given by the formula:
$n = \frac{\text{Volume at STP}}{\text{Molar Volume at STP}}$
$n = \frac{0.224 \ L}{22.4 \ L \ mol^{-1}} = 0.01 \ mol$.

(A) Since the temperature is constant,we use Boyle's Law: $P_1V_1 = P_2V_2$.
Given: $P_1 = 730 \, mm$,$V_1 = 380 \, mL$,$P_2 = 760 \, mm$.
Substituting the values: $V_2 = \frac{P_1 \times V_1}{P_2} = \frac{730 \times 380}{760} = 365 \, mL$.

(D) The value of the gas constant $R$ depends on the units used for pressure,volume,and temperature.

(D) According to Charles's Law,$V \propto T$. When air is heated,it expands and its density decreases,making the hot air lighter.

(A) According to Boyle's Law,$V \propto \frac{1}{P}$ at constant temperature.
Since the pressure of the atmosphere is higher at sea level,the volume occupied by a given mass of air decreases,leading to a higher density of air.

(C) From the ideal gas equation,$PV = nRT$,the molar volume $V_m = \frac{V}{n} = \frac{RT}{P}$.
Thus,$V_m \propto \frac{T}{P}$.
For maximum molar volume,we need the highest temperature $(T)$ and the lowest pressure $(P)$.
Comparing the conditions:
$A$: $T = 273.15\, K, P = 1\, atm \implies \frac{T}{P} = 273.15$
$B$: $T = 273.15\, K, P = 2\, atm \implies \frac{T}{P} = 136.57$
$C$: $T = 127 + 273 = 400\, K, P = 1\, atm \implies \frac{T}{P} = 400$
$D$: $T = 273 + 273 = 546\, K, P = 2\, atm \implies \frac{T}{P} = 273$
Comparing the values,the ratio $\frac{T}{P}$ is maximum for option $C$.

(A) According to Boyle's Law,for a fixed amount of gas at a constant temperature,the product of pressure $(P)$ and volume $(V)$ is constant $(PV = k)$.
Therefore,a plot of $PV$ versus $P$ results in a straight line parallel to the $P$-axis ($X$-axis).

(B) Using the ideal gas equation $PV = nRT$,we have $n = \frac{PV}{RT}$.
For vessel $A$: $n_A = \frac{P_A V_A}{R T_A}$.
For vessel $B$: $n_B = \frac{P_B V_B}{R T_B}$.
Given: $P_A = 2P_B$,$V_A = 2V_B$,and $T_A = 2T_B$.
Substituting these values: $\frac{n_A}{n_B} = \frac{(2P_B)(2V_B) / (R \times 2T_B)}{(P_B V_B) / (R T_B)} = \frac{2 P_B V_B / R T_B}{P_B V_B / R T_B} = 2$.
Therefore,the ratio of molecules in $A$ to $B$ is $2:1$.

(A) Using the ideal gas law equation $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Given: $P_1 = P$,$P_2 = 2P$,$V_1 = V$,$V_2 = 0.85 V$ (since volume reduced by $15\%$).
Initial temperature $T_1 = 75 + 273.15 = 348.15 \, K$.
Substituting the values: $\frac{P \times V}{348.15} = \frac{2P \times 0.85 V}{T_2}$.
$T_2 = \frac{2 \times 0.85 \times 348.15}{1} = 591.855 \, K$.
Converting back to Celsius: $T_2(^\circ C) = 591.855 - 273.15 = 318.705 \, ^\circ C \approx 319 \, ^\circ C$.

(B) Using the ideal gas equation $PV = nRT = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass of $CH_4$ $(16 \ g/mol)$.
Given: $P = 1 \ atm$,$V = 9 \ L$,$T = 27 + 273 = 300 \ K$,$R = 0.0821 \ L \ atm \ K^{-1} mol^{-1}$.
Rearranging for mass $m$: $m = \frac{PVM}{RT}$.
Substituting the values: $m = \frac{1 \times 9 \times 16}{0.0821 \times 300}$.
$m = \frac{144}{24.63} \approx 5.85 \ g$.

(A) Using the ideal gas equation $PV = \frac{m}{M}RT$,where $M$ is the molar mass.
Given: $P = 1 \, atm$,$V = 4.1 \, L$,$m = 7.0 \, g$,$T = 300 \, K$,and $R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.
Rearranging for $M$: $M = \frac{mRT}{PV}$.
Substituting the values: $M = \frac{7.0 \times 0.0821 \times 300}{1 \times 4.1}$.
$M = \frac{172.41}{4.1} \approx 42.05 \, g/mol$.
Thus,the molar mass is approximately $42 \, g/mol$.

(B) Since the pressure and volume of the $H_2$ gas and the given gas are the same,we use the ideal gas equation $PV = nRT = \frac{m}{M}RT$.
For the given gas: $P V = \frac{3.7}{M} R (25 + 273) = \frac{3.7}{M} R (298)$.
For $H_2$ gas: $P V = \frac{0.184}{2} R (17 + 273) = \frac{0.184}{2} R (290)$.
Equating both expressions: $\frac{3.7}{M} \times 298 = \frac{0.184}{2} \times 290$.
Solving for $M$: $M = \frac{3.7 \times 298 \times 2}{0.184 \times 290} \approx 41.3 \, g \, mol^{-1}$.

(A) Using the ideal gas equation $PV = nRT = \frac{m}{M}RT$,
Given: $V = 600 \, mL = 0.6 \, L$,$m = 22 \, g$,$M = 44 \, g/mol$,$R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,$T = 25 + 273 = 298 \, K$.
Calculating the pressure exerted by $CO_2$ gas:
$P = \frac{m \times R \times T}{M \times V} = \frac{22 \times 0.0821 \times 298}{44 \times 0.6} = 0.5 \times \frac{24.4658}{0.6} = 20.388 \approx 20.4 \, atm$.
Since the container was initially empty (vacuum) or the question implies the total pressure inside the sealed vessel after evaporation,the pressure is $20.4 \, atm$.

(C) The decrease in the weight of water is due to the water molecules being carried away by the $N_2$ gas. This water vapor occupies the same volume as the $N_2$ gas,which is $48 \, L$.
Using the ideal gas equation $PV = \frac{m}{M}RT$,where $V = 48 \, L$,$m = 1.2 \, g$,$M = 18 \, g/mol$,$R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,and $T = 27 + 273 = 300 \, K$.
$P = \frac{mRT}{MV} = \frac{1.2 \times 0.0821 \times 300}{18 \times 48}$
$P = \frac{29.556}{864} = 0.034 \, atm$.

(A) Using Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = 1 \ L = 1000 \ mL$,$T_1 = 25 + 273 = 298 \ K$,$T_2 = 35 + 273 = 308 \ K$.
Calculating $V_2$: $V_2 = \frac{V_1 \times T_2}{T_1} = \frac{1000 \times 308}{298} \approx 1033.56 \ mL$.
The volume of air expelled = $V_2 - V_1 = 1033.56 - 1000 = 33.56 \ mL \approx 33 \ mL$.

(B) The graph shows the relationship between volume $(V)$ and temperature $(T)$ at constant pressure.
According to Charles's Law,the volume of a fixed mass of gas is directly proportional to its absolute temperature at constant pressure $(V \propto T)$.
The plot of $V$ versus $T$ $(^{\circ}C)$ is a straight line that intersects the temperature axis at $-273.15 \ ^{\circ}C$ (absolute zero).

(D) Using the principle of conservation of moles (Boyle's Law for mixtures): $P_1V_1 + P_2V_2 = P_3(V_1 + V_2)$
Given: $P_1 = 100 \ kPa, V_1 = 1 \ L$
$P_2 = 320 \ kPa, V_2 = 3 \ L$
Total volume $V_{total} = V_1 + V_2 = 1 + 3 = 4 \ L$
Substituting the values: $P_3(4) = (100 \times 1) + (320 \times 3)$
$P_3(4) = 100 + 960 = 1060$
$P_3 = \frac{1060}{4} = 265 \ kPa$

(A) Using Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ (at constant pressure).
Given: $V_1 = 0.2 \, L$,$T_1 = 0 \, ^oC = 273 \, K$,$T_2 = 273 \, ^oC = 273 + 273 = 546 \, K$.
Substituting the values: $\frac{0.2}{273} = \frac{V_2}{546}$.
$V_2 = \frac{0.2 \times 546}{273} = 0.2 \times 2 = 0.4 \, L$.

(C) According to Boyle's Law,for a fixed amount of gas at a constant temperature,$PV = k$ (a constant).
Therefore,the product $PV$ remains constant regardless of the change in pressure $P$.
This results in a horizontal line in a graph of $PV$ versus $P$,which means the slope of the graph is zero.

(D) Charles's Law states that for a fixed mass of an ideal gas at constant pressure,the volume is directly proportional to its absolute temperature $(V \propto T)$.
Mathematically,$V_t = V_0 \left( 1 + \frac{t}{273.15} \right) = V_0 \left( \frac{273.15 + t}{273.15} \right)$.
This can be written as $V_t = \left( \frac{V_0}{273.15} \right) T$,where $T = 273.15 + t$.
This equation is of the form $V_t = mT$,which is a linear equation $V_t = a + bt$ (where $a=0$ and $b = V_0/273.15$).
Option $D$,$V_t = V_0 t$,does not represent this linear relationship.

(A) The ideal gas equation is given by $PV = nRT$.
Since $n = \frac{m}{M}$ (where $m$ is mass and $M$ is molar mass),we have $PV = \frac{m}{M}RT$.
Rearranging gives $PM = \frac{m}{V}RT$.
Since density $d = \frac{m}{V}$,the equation becomes $PM = dRT$.

(D) We know that the gas constant $R = 8.314 \, J \, K^{-1} mol^{-1} = 8.314 \, N \cdot m \cdot K^{-1} mol^{-1}$.
Since $1 \, N = 10^5 \, dyne$ and $1 \, m = 10^3 \, mm$,we have $1 \, m^3 = (10^3 \, mm)^3 = 10^9 \, mm^3$.
However,the pressure unit given is $dyne \, m^{-2}$. Note that $1 \, N \cdot m = 10^5 \, dyne \cdot m$.
To convert $R$ to the required units:
$R = 8.314 \, N \cdot m \cdot K^{-1} mol^{-1} = 8.314 \times (10^5 \, dyne) \times (10^9 \, mm^3) \cdot m^{-2} \cdot K^{-1} mol^{-1}$ is incorrect based on the unit provided.
Correct conversion: $1 \, J = 10^7 \, erg = 10^7 \, dyne \cdot cm$.
Given pressure $P$ in $dyne \cdot m^{-2}$ and volume $V$ in $mm^3$:
$R = 8.314 \, J = 8.314 \times 10^7 \, dyne \cdot cm = 8.314 \times 10^7 \times (10^{-2} \, m) \times (10^3 \, mm)^3 \times (10^{-6} \, m^2)^{-1} \dots$
Actually,$1 \, J = 10^7 \, dyne \cdot cm = 10^7 \, dyne \cdot (10^{-2} \, m) = 10^5 \, dyne \cdot m$.
$1 \, m^3 = 10^9 \, mm^3$,so $1 \, m = 10^3 \, mm$.
$R = 8.314 \, N \cdot m = 8.314 \times (10^5 \, dyne) \times (10^9 \, mm^3) \times (10^{-3} \, m^{-1}) \dots$
Simplifying: $R = 8.314 \times 10^5 \, dyne \cdot m \cdot K^{-1} mol^{-1} = 8.314 \times 10^5 \times (10^9 \, mm^3) \times (10^{-2} \, m^{-2}) = 8.314 \times 10^{14} \, (dyne \cdot m^{-2})(mm^3) K^{-1} mol^{-1}$.

(A) From the ideal gas equation,$PV = nRT = \frac{m}{M} RT$.
Since density $d = \frac{m}{V}$,we have $P = \frac{dRT}{M}$,which implies $d = \frac{PM}{RT}$.
For a constant mass of gas,$\frac{d_1 T_1}{P_1} = \frac{d_2 T_2}{P_2}$.
Rearranging for $d_2$,we get $d_2 = d_1 \left( \frac{T_1 P_2}{T_2 P_1} \right)$.

(A) Using the ideal gas law relation for density: $\frac{d_1 T_1}{P_1} = \frac{d_2 T_2}{P_2}$
At $STP$,$d_1 = 1.43 \, g \, L^{-1}$,$T_1 = 273 \, K$,$P_1 = 760 \, torr$.
At the given conditions,$T_2 = 27 + 273 = 300 \, K$,$P_2 = 700 \, torr$.
Rearranging for $d_2$: $d_2 = \frac{d_1 \times T_1 \times P_2}{P_1 \times T_2}$
$d_2 = \frac{1.43 \times 273 \times 700}{760 \times 300} \approx 1.20 \, g \, L^{-1}$.

(C) From the ideal gas equation,$PV = nRT$,we know that $PM = dRT$,where $P$ is pressure,$M$ is molar mass,$d$ is density,$R$ is the gas constant,and $T$ is temperature.
For gas $x$: $P_x M_x = d_x RT$
For gas $y$: $P_y M_y = d_y RT$
Dividing the two equations: $\frac{P_x M_x}{P_y M_y} = \frac{d_x}{d_y}$
Given: $d_x = 2d_y$ and $M_x = \frac{1}{3} M_y$
Substituting these values: $\frac{P_x}{P_y} \times \frac{1}{3} = \frac{2d_y}{d_y} = 2$
Therefore,$\frac{P_x}{P_y} = 2 \times 3 = 6$.

(B) From the ideal gas equation,$PV = nRT$,we have $n = \frac{PV}{RT}$.
For container $X$: $n_X = \frac{P_X V_X}{R T_X}$.
For container $Y$: $n_Y = \frac{P_Y V_Y}{R T_Y}$.
Given $P_X = 3P_Y$,$V_X = 3V_Y$,and $T_X = 3T_Y$.
Substituting these values: $n_X = \frac{(3P_Y)(3V_Y)}{R(3T_Y)} = 3 \times \frac{P_Y V_Y}{R T_Y} = 3n_Y$.
Since $n = \frac{\text{mass}}{\text{molar mass}}$,and the gas is the same,the molar mass is constant.
Therefore,$\frac{m_X}{M} = 3 \times \frac{m_Y}{M}$.
Given $m_X = m$,we get $m = 3m_Y$.
Thus,$m_Y = \frac{m}{3} \, g$.

(A) Using the ideal gas equation: $PV = nRT$,where $n = \frac{w}{M}$.
So,$PV = \frac{w}{M}RT$,which implies $\frac{PV}{wT} = \frac{R}{M} = \text{constant}$.
Therefore,$\frac{P_1 V_1}{w_1 T_1} = \frac{P_2 V_2}{w_2 T_2}$.
Given: $P_1 = 1 \ atm, V_1 = 1 \ L, w_1 = 2 \ g, T_1 = 300 \ K$.
Given: $P_2 = 0.75 \ atm, V_2 = 1 \ L, w_2 = 1 \ g, T_2 = ?$.
Substituting the values: $\frac{1 \times 1}{2 \times 300} = \frac{0.75 \times 1}{1 \times T_2}$.
$T_2 = 0.75 \times 2 \times 300 = 450 \ K$.

(B) According to the ideal gas equation,$PV = nRT$,which can be rearranged as $V = (nR/P)T$.
This represents a straight line equation $y = mx$,where the slope $m = nR/P$.
Since the slope is inversely proportional to pressure $(m \propto 1/P)$,a smaller slope corresponds to a higher pressure.
In the given graph,the slope of the line for $P_2$ is smaller than the slope of the line for $P_1$.
Therefore,$P_2 > P_1$ or $P_1 < P_2$.

(A) From the ideal gas equation $PV = nRT$,we know that $P = \frac{dRT}{M}$.
Assuming the molar masses $M$ are the same or comparing the ratio of pressures for the same gas under different conditions,we use the relation $\frac{P_1 M_1}{d_1 T_1} = \frac{P_2 M_2}{d_2 T_2}$.
Given $\frac{d_1}{d_2} = \frac{1}{2}$ and $\frac{T_1}{T_2} = \frac{2}{1}$.
Assuming $M_1 = M_2$,the ratio of pressures is $\frac{P_1}{P_2} = \frac{d_1 T_1}{d_2 T_2}$.
Substituting the values: $\frac{P_1}{P_2} = (\frac{1}{2}) \times (\frac{2}{1}) = \frac{1}{1}$.
Thus,the ratio of pressures is $1:1$.

(B) $1$ mole of $CO$ has a molar mass of $28 \, g \, mol^{-1}$.
At $STP$,$1$ mole of any ideal gas occupies $22.4 \, L$.
Density $(\rho)$ = $\frac{\text{Mass}}{\text{Volume}} = \frac{28 \, g}{22.4 \, L} = 1.25 \, g \, L^{-1}$.

(B) Using the ideal gas equation $PV = nRT$,where $n = \frac{m}{M}$.
Density $d = \frac{m}{V} = \frac{PM}{RT}$.
For ammonia $(NH_3)$,molar mass $M = 14 + 3(1) = 17 \, g \, mol^{-1}$.
Given $P = 3 \, atm$,$T = 27 + 273 = 300 \, K$,and $R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$.
$d = \frac{3 \times 17}{0.0821 \times 300} = \frac{51}{24.63} \approx 2.07 \, g \, L^{-1}$.
Since $1 \, L = 1 \, dm^3$,the density is $2.07 \, g \, dm^{-3}$.

(C) According to the ideal gas law,$PV = nRT$.
Since the temperature $(T)$ and the volume of the container $(V)$ are constant,the pressure $(P)$ is directly proportional to the number of moles $(n)$ of the gas $(P \propto n)$.
As the gas is continuously introduced into the evacuated container,the number of moles $(n)$ increases,causing the pressure $(P)$ to increase.
Once the container is filled to its capacity or the gas supply is stopped,the number of moles becomes constant,and consequently,the pressure inside the container becomes constant.

(A) The ideal gas equation is given by $PV = nRT$.
From this equation,we can express the relationship between variables as $n = \frac{PV}{RT}$.
$(a)$ For $n \text{ vs } V$,at constant $P$ and $T$,$n \propto V$,which is a straight line.
$(b)$ For $T \text{ vs } P$,at constant $n$ and $V$,$T \propto P$,which is a straight line.
$(c)$ For $n \text{ vs } \frac{1}{T}$,at constant $P$ and $V$,$n \propto \frac{1}{T}$,which is a straight line.
$(d)$ For $n \text{ vs } \frac{1}{P}$,at constant $V$ and $T$,$n \propto \frac{1}{P}$,which is a straight line.
However,if we consider the standard interpretation of such problems where variables are plotted against each other,all these represent linear relationships under specific constant conditions. If the question implies a general case where variables are not held constant,none would be straight lines. Given the options,all represent linear relationships $y = mx$ under specific constraints.

(B) Using the ideal gas equation $PV = nRT = \frac{w}{M}RT$,
Given:
$P = 0.821 \ atm$
$w = 2.8 \ g$
$M_{CO} = 28 \ g/mol$
$R = 0.0821 \ L \ atm \ mol^{-1} K^{-1}$
$T = 27 \ ^oC = 27 + 273 = 300 \ K$
Substituting the values:
$V = \frac{wRT}{PM} = \frac{2.8 \times 0.0821 \times 300}{0.821 \times 28}$
$V = \frac{2.8}{28} \times \frac{0.0821 \times 300}{0.821}$
$V = 0.1 \times 10 \times 3 = 3 \ L$

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

(D) Using the ideal gas law equation: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Given: $P_1 = P, P_2 = 2P, V_1 = 4 \, dm^3, V_2 = ?$
Given: $T_1 = T, T_2 = 2T$.
Substituting the values: $\frac{P \times 4}{T} = \frac{2P \times V_2}{2T}$.
Solving for $V_2$: $V_2 = \frac{P \times 4 \times 2T}{T \times 2P} = 4 \, dm^3$.

(B) Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
Given: $P_1 = 1 \, atm$,$V_1 = 100 \, cc$,$T_1 = 0 \, ^oC = 273 \, K$
New conditions: $P_2 = 1.5 \, P_1 = 1.5 \, atm$
$T_2 = T_1 + \frac{1}{3} T_1 = 273 + \frac{273}{3} = 273 + 91 = 364 \, K$
Substituting the values: $V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} = \frac{1 \times 100 \times 364}{273 \times 1.5}$
$V_2 = \frac{36400}{409.5} \approx 88.9 \, cc$

(C) We know the ideal gas equation is $PV = nRT$.
Rearranging the equation to find the number of moles per unit volume (moles per liter),we get:
$n/V = P/(RT)$.

(D) Total moles $n_{total} = n_{N_2} + n_{He} + n_{O_2} = 0.3 + 0.5 + 6.2 = 7.0 \, mol$.
Using the ideal gas equation $PV = nRT$,where $T = 27 + 273 = 300 \, K$ and $R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$:
$P_{total} = \frac{n_{total}RT}{V} = \frac{7.0 \times 0.0821 \times 300}{2.461} = \frac{172.41}{2.461} \approx 70 \, atm$.
The partial pressure of $N_2$ is given by $P_{N_2} = \chi_{N_2} \times P_{total}$,where $\chi_{N_2}$ is the mole fraction of $N_2$.
$\chi_{N_2} = \frac{n_{N_2}}{n_{total}} = \frac{0.3}{7.0}$.
$P_{N_2} = \frac{0.3}{7.0} \times 70 = 3 \, atm$.

(A) Using the ideal gas equation $PV = nRT = \frac{w}{M}RT$ for both gases separately:
For $O_2$: $n_{O_2} = \frac{4\, g}{32\, g/mol} = 0.125\, mol$.
$P_{O_2} = \frac{n_{O_2}RT}{V} = \frac{0.125 \times 0.0821 \times 273}{1} = 2.801\, atm$.
For $H_2$: $n_{H_2} = \frac{2\, g}{2\, g/mol} = 1.0\, mol$.
$P_{H_2} = \frac{n_{H_2}RT}{V} = \frac{1.0 \times 0.0821 \times 273}{1} = 22.413\, atm$.
According to Dalton's Law of Partial Pressures,$P_{total} = P_{O_2} + P_{H_2} = 2.801 + 22.413 = 25.214\, atm$.

(B) At $27 \, ^oC$,the pressure of the dry gas is calculated as:
$P_{dry} = P_{total} - P_{water} = 725 \, torr - 25 \, torr = 700 \, torr$.
Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
At $STP$: $P_1 = 760 \, torr$,$V_1 = 136.5 \, cm^3$,$T_1 = 273 \, K$.
At the given conditions: $P_2 = 700 \, torr$,$T_2 = 27 + 273 = 300 \, K$.
Substituting the values: $V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} = \frac{760 \times 136.5 \times 300}{273 \times 700} = 162.9 \, cm^3$.

(B) For an ideal gas,the ideal gas equation is $PV = nRT$.
The compressibility factor $Z$ is defined as $Z = \frac{PV}{nRT}$.
Since $PV = nRT$ for an ideal gas,substituting this into the equation gives $Z = \frac{nRT}{nRT} = 1$.

(B) For an ideal gas,the intermolecular forces of attraction are considered to be zero.
During free expansion into a vacuum,the external pressure is zero,so no work is done $(w = 0)$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Since the process is adiabatic $(q = 0)$ and $w = 0$,the change in internal energy $\Delta U = 0$.
For an ideal gas,internal energy depends only on temperature $(U = f(T))$.
Therefore,$\Delta U = 0$ implies $\Delta T = 0$.
Thus,no cooling or heating occurs.

(C) The ideal gas equation $PV = nRT$ is a state function equation that describes the relationship between pressure $(P)$,volume $(V)$,amount of substance $(n)$,and temperature $(T)$.
It is applicable to any process involving an ideal gas,regardless of whether the process is isothermal,adiabatic,isobaric,or isochoric.

(B) The deviation of a gas from ideal behavior is expressed by the compressibility factor,denoted as $Z$.
For an ideal gas,$PV = nRT$,so $Z = 1$.
For real gases,the compressibility factor is defined as $Z = \frac{PV}{nRT}$.