Gas Laws (Charles, Boyle's, Avagadro's, Gay Lussacs and Dalton's law) and Ideal gas Equation Questions in English

(C) For an ideal gas,the equation of state is given by $PV = nRT$.
Rearranging this equation,we get $\frac{PV}{T} = nR$.
Since $n$ (number of moles) and $R$ (universal gas constant) are constants for a given amount of gas,the ratio $\frac{PV}{T}$ remains constant.
This relationship is derived from the ideal gas law and is valid for any process involving an ideal gas,including isothermal and adiabatic processes.
Therefore,option $C$ is correct.

(C) For an isobaric process (constant pressure),the ideal gas law $PV = nRT$ implies that $V \propto T$ (Charles's Law).
Given $n = 1$ mole,initial volume is $V$ at temperature $T$.
When temperature increases by $1 \, K$,the new temperature is $T' = T + 1$.
The new volume $V'$ is given by $\frac{V'}{T+1} = \frac{V}{T}$.
$V' = V \left(1 + \frac{1}{T}\right) = V + \frac{V}{T}$.
For a gas at standard conditions or generally,the coefficient of volume expansion $\gamma = \frac{1}{T}$. For an ideal gas,$\gamma = \frac{1}{273} \, K^{-1}$ at $0 \, ^\circ C$ $(273 \, K)$.
Thus,the increase in volume $\Delta V = V' - V = \frac{V}{T}$.
Assuming $T = 273 \, K$,the increase in volume is $\frac{V}{273}$.

(C) According to Charles's Law,for a fixed mass of gas at constant pressure,the volume is directly proportional to the absolute temperature: ${V \propto T}$ or $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: Initial volume ${V_1 = V_0}$,Initial temperature ${T_1 = 27^{\circ}C = 27 + 273 = 300 K}$.
Final volume ${V_2 = 2V_0}$.
Substituting the values: $\frac{V_0}{300} = \frac{2V_0}{T_2}$.
Solving for ${T_2}$: ${T_2 = 2 \times 300 = 600 K}$.
Converting back to Celsius: ${T_2 = 600 - 273 = 327^{\circ}C}$.

(C) According to Charles's Law,at constant pressure,the volume of a given mass of gas is directly proportional to its absolute temperature: $V \propto T$.
This implies $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given:
$V_1 = 300 \ ml$
$T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$
$T_2 = 7^{\circ}C = 7 + 273 = 280 \ K$
Substituting the values into the formula:
$V_2 = \frac{V_1 \times T_2}{T_1} = \frac{300 \times 280}{300} = 280 \ ml$.

(A) For an isothermal process, $PV = \text{constant}$.
Differentiating both sides with respect to $P$, we get:
$P(dV/dP) + V = 0$
$P(dV/dP) = -V$
$-(dV/dP)/V = 1/P$
Given $\beta = -(dV/dP)/V$, we have $\beta = 1/P$.
This equation represents a rectangular hyperbola, where $\beta$ decreases as $P$ increases. This relationship is correctly shown in graph $A$.

(A) For an isothermal process,Boyle's Law states that $P_1 V_1 = P_2 V_2$.
Given: $V_1 = 1 \, L = 1000 \, cm^3$,$P_1 = 72 \, cm$ of $Hg$,$V_2 = 900 \, cm^3$.
Using the formula: $P_2 = \frac{P_1 V_1}{V_2} = \frac{72 \times 1000}{900} = 80 \, cm$ of $Hg$.
The change in pressure (stress) is $\Delta P = P_2 - P_1$.
$\Delta P = 80 \, cm - 72 \, cm = 8 \, cm$ of $Hg$.

(C) From the ideal gas equation,$PV = \mu RT$,we can write $V = \frac{\mu RT}{P}$,which implies $V \propto \frac{T}{P}$.
In the given $P-T$ graph,the line representing the process from $1$ to $2$ passes through the $T$-axis at a positive value (if extended). This means the slope $\frac{P}{T}$ is decreasing as $T$ increases.
Since $V \propto \frac{T}{P}$,if the ratio $\frac{P}{T}$ decreases,the ratio $\frac{T}{P}$ must increase.
Therefore,the volume $V$ of the gas increases during the process from $1$ to $2$.

(C) For an ideal gas,the equation of state is $PV = nRT$.
This implies $P = \frac{nRT}{V}$.
For a constant volume $V$,the pressure $P$ is directly proportional to the temperature $T$ $(P \propto T)$.
In the given $P-V$ diagram,if we draw a vertical line at a constant volume $V$,the pressure corresponding to the curve $T_2$ is higher than the pressure corresponding to the curve $T_1$.
Since $P_2 > P_1$ at the same volume,it follows that $T_2 > T_1$,or $T_1 < T_2$.

(C) According to the ideal gas equation,$PV = nRT$.
For a constant pressure $P$ and a fixed amount of gas $n$,the equation becomes $V = (\frac{nR}{P})T$.
This is of the form $V = kT$,where $k = \frac{nR}{P}$ is a constant.
This represents a straight line passing through the origin $(0, 0)$ in the $V-T$ plane.
Therefore,the correct graph is a straight line passing through the origin.

(B) The ideal gas equation is given by $PV = nRT$,where $n$ is the number of moles.
$n = \frac{N}{N_A}$,where $N$ is the total number of molecules and $N_A$ is Avogadro's number.
Also,the total mass of the gas is $M_{total} = N \times m$,where $m$ is the mass of one molecule.
Density $\rho = \frac{M_{total}}{V} = \frac{Nm}{V}$.
From the ideal gas law: $PV = \frac{N}{N_A} RT$.
Since $R = k N_A$,we have $PV = \frac{N}{N_A} (k N_A) T = NkT$.
Rearranging for $N/V$: $\frac{N}{V} = \frac{P}{kT}$.
Substituting this into the density formula: $\rho = \left(\frac{N}{V}\right) m = \left(\frac{P}{kT}\right) m = \frac{Pm}{kT}$.

(A) According to the Ideal Gas Law,$PV = NkT$,where $P$ is pressure,$V$ is volume,$N$ is the number of molecules,$k$ is the Boltzmann constant,and $T$ is temperature.
Rearranging for the number of molecules,we get $N = \frac{PV}{kT}$.
Since $P$,$V$,$k$,and $T$ are identical for both gases,the number of molecules $N$ must be the same for both gases.

(A) According to the ideal gas equation,$PV = nRT$,where $n$ is the number of moles.
Given that $P, V, R$,and $T$ are the same for all three gases,the number of moles $n = \frac{PV}{RT}$ is the same for all.
Number of molecules $N = n \times N_A$,where $N_A$ is Avogadro's number. Thus,the number of molecules is the same for all.
Now,we calculate the number of atoms:
For $H_2$ (diatomic): Number of atoms $= 2 \times n \times N_A = 2nN_A$.
For $O_2$ (diatomic): Number of atoms $= 2 \times n \times N_A = 2nN_A$.
For $He$ (monoatomic): Number of atoms $= 1 \times n \times N_A = nN_A$.
Comparing these,$H_2$ and $O_2$ have the same number of atoms,which is higher than that of $He$.

(C) Using the ideal gas law equation: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Given: $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$,$P_1 = P$,$V_1 = V$.
Final conditions: $P_2 = 2P$,$V_2 = 3V$.
Substituting these values into the equation:
$\frac{P \times V}{300} = \frac{(2P) \times (3V)}{T_2}$.
$T_2 = \frac{6PV}{PV} \times 300 = 6 \times 300 = 1800 \ K$.
To convert to Celsius: $T(^{\circ}C) = 1800 - 273 = 1527^{\circ}C$.

(D) According to Boyle's Law for a fixed amount of gas at constant temperature,the product of pressure and volume remains constant: $P_1V_1 = P_2V_2$.
Given values are $P_1 = 3 \, atm$,$V_1 = 12 \, L$,and $V_2 = 9 \, L$.
Substituting these values into the equation: $3 \times 12 = P_2 \times 9$.
Solving for $P_2$: $P_2 = \frac{3 \times 12}{9} = \frac{36}{9} = 4 \, atm$.
Therefore,the required pressure is $4 \, atm$.

(D) Using the ideal gas equation $PV = Nk_BT$,where $N$ is the number of molecules and $k_B$ is the Boltzmann constant.
For container $A$: $N_A = \frac{PV}{k_BT}$.
For container $B$: $N_B = \frac{(2P)(V/4)}{k_B(2T)} = \frac{PV/2}{2k_BT} = \frac{PV}{4k_BT}$.
The ratio of the number of molecules is $\frac{N_A}{N_B} = \frac{PV/k_BT}{PV/4k_BT} = \frac{1}{1/4} = \frac{4}{1}$.

(D) The density $\rho$ of an ideal gas is given by the formula $\rho = \frac{M_w P}{RT}$,where $M_w$ is the molar mass,$P$ is the pressure,$R$ is the gas constant,and $T$ is the absolute temperature.
Given that the pressure $P' = 2P$ and the temperature $T' = 4T$.
The new density $\rho'$ is given by $\rho' = \frac{M_w P'}{R T'}$.
Substituting the values,we get $\rho' = \frac{M_w (2P)}{R (4T)} = \frac{1}{2} \left( \frac{M_w P}{RT} \right)$.
Therefore,$\rho' = 0.5 \rho$.
The density becomes $0.5$ times the original density.

(C) Since the two spheres are connected,the pressure $P$ in both spheres is the same.
Given: Volume $V_1 = 2V_2$,Temperature $T_1 = 100 K$,$T_2 = 200 K$.
Using the ideal gas equation $PV = nRT = (m/M)RT$,where $M$ is the molar mass of the gas.
For sphere $I$: $PV_1 = (m/M)RT_1$ --- $(1)$
For sphere $II$: $PV_2 = (m'/M)RT_2$ --- $(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{PV_2}{PV_1} = \frac{(m'/M)RT_2}{(m/M)RT_1}$
$\frac{V_2}{2V_2} = \frac{m'}{m} \times \frac{200}{100}$
$\frac{1}{2} = \frac{m'}{m} \times 2$
$\frac{m'}{m} = \frac{1}{4}$
$m' = m/4$

(B) Since the temperature is constant,we use Boyle's Law: $P_1 V_1 = P_2 V_2$.
Given: $P_1 = 2 \, atm$,$V_1 = 500 \, cc$,$V_2 = 400 \, cc$.
Substituting the values into the equation:
$P_2 = P_1 \times \frac{V_1}{V_2}$
$P_2 = 2 \times \frac{500}{400}$
$P_2 = 2 \times 1.25 = 2.5 \, atm$.

(B) According to Boyle's Law,at constant temperature,$P_1V_1 = P_2V_2$.
Here,the initial pressure $P_1 = 0.8 \, m$ and initial volume $V_1 = 5 \, L$.
The final volume $V_2$ is the sum of the two vessels because the gas expands to fill both: $V_2 = 5 \, L + 3 \, L = 8 \, L$.
Substituting the values into the equation:
$0.8 \times 5 = P_2 \times 8$
$4.0 = P_2 \times 8$
$P_2 = 4.0 / 8 = 0.5 \, m$.
Thus,the resultant pressure is $0.5 \, m$.

(D) According to Gay-Lussac's Law,for a given mass of gas at constant volume,the pressure is directly proportional to its absolute temperature: $P \propto T$.
Therefore,$\frac{P_2}{P_1} = \frac{T_2}{T_1}$.
Given: $P_1 = P$,$P_2 = 2P$,and $T_1 = 20^{\circ}C = 20 + 273 = 293 \ K$.
Substituting the values: $\frac{2P}{P} = \frac{T_2}{293}$.
$T_2 = 2 \times 293 = 586 \ K$.
To convert the temperature back to Celsius: $T(^{\circ}C) = 586 - 273 = 313^{\circ}C$.

(D) The ideal gas equation is given by $PV = NkT$,where $N$ is the number of molecules and $k$ is the Boltzmann constant.
For container $A$: $P_A = P$,$V_A = V$,$T_A = T$. So,$N_A = \frac{PV}{kT}$.
For container $B$: $P_B = 2P$,$V_B = V/4$,$T_B = 2T$. So,$N_B = \frac{(2P)(V/4)}{k(2T)} = \frac{PV/2}{2kT} = \frac{PV}{4kT}$.
The ratio of the number of molecules is $\frac{N_A}{N_B} = \frac{PV/kT}{PV/4kT} = \frac{4}{1}$.
Therefore,the ratio is $4 : 1$.

(B) Using the ideal gas equation,$PV = \frac{m}{M}RT$,where $m$ is the mass of the gas,$M$ is the molar mass,$R$ is the gas constant,and $T$ is the temperature.
Since the volume $V$ and molar mass $M$ are constant,we have $P \propto \frac{mT}{V} \implies P \propto mT$.
Let $m_1 = 6 \ g$,$P_1 = P$,$T_1 = 400 \ K$ and $m_2$ be the final mass,$P_2 = P/2$,$T_2 = 300 \ K$.
Using the ratio $\frac{P_1}{P_2} = \frac{m_1 T_1}{m_2 T_2}$,we get $\frac{P}{P/2} = \frac{6 \times 400}{m_2 \times 300}$.
$2 = \frac{2400}{300 m_2} \implies 2 = \frac{8}{m_2}$.
$m_2 = 4 \ g$.
The amount of oxygen leaked is $\Delta m = m_1 - m_2 = 6 \ g - 4 \ g = 2 \ g$.

(D) According to Boyle's Law,for a given mass of gas at constant temperature,$P_1V_1 = P_2V_2$.
Let the initial pressure be $P_1 = P$ and initial volume be $V_1 = V$.
The final volume is $V_2 = V - 0.1V = 0.9V$.
Using $P_1V_1 = P_2V_2$,we get $P \times V = P_2 \times 0.9V$.
Therefore,$P_2 = \frac{P}{0.9} = \frac{10}{9}P$.
The change in pressure is $\Delta P = P_2 - P = \frac{10}{9}P - P = \frac{1}{9}P$.
The percentage increase in pressure is $\frac{\Delta P}{P} \times 100 = \frac{1/9P}{P} \times 100 = \frac{100}{9} \approx 11.11\%$.

(D) Using the ideal gas equation $PV = nRT$,we can write $P = (n/V)RT$,where $n/V$ is the number density of molecules.
Given $P = nkT$,where $n$ is the number density and $k$ is the Boltzmann constant $(k = 1.38 \times 10^{-23} \; J/K)$.
First,convert pressure to $SI$ units $(Pa)$:
$1 \; atm = 760 \; mm \; Hg = 1.013 \times 10^{5} \; Pa$.
$P = 10^{-11} \; mm \; Hg = (10^{-11} / 760) \times 1.013 \times 10^{5} \; Pa \approx 1.33 \times 10^{-9} \; Pa$.
Temperature $T = 27 + 273 = 300 \; K$.
Using $n = P / (kT)$:
$n = (1.33 \times 10^{-9}) / (1.38 \times 10^{-23} \times 300) \approx 3.21 \times 10^{11} \; molecules/m^{3}$.
Since $1 \; m^{3} = 10^{6} \; cm^{3}$,the number density per $cm^{3}$ is $3.21 \times 10^{11} / 10^{6} = 3.21 \times 10^{5} \; molecules/cm^{3}$.
Rounding to the nearest option,the answer is $3.22 \times 10^{5} \; molecules/cm^{3}$.

(C) For a gas in a closed vessel,the volume $V$ is constant. According to Gay-Lussac's Law,$\frac{P}{T} = \text{constant}$,which means $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Let the initial pressure be $P_1 = P$ and the initial temperature be $T_1 = T$ (in Kelvin).
The final pressure is $P_2 = P + 0.4\% \text{ of } P = P + 0.004P = 1.004P$.
The final temperature is $T_2 = T + 1$.
Substituting these values into the equation: $\frac{P}{T} = \frac{1.004P}{T+1}$.
Dividing both sides by $P$: $\frac{1}{T} = \frac{1.004}{T+1}$.
Cross-multiplying: $T + 1 = 1.004T$.
$1 = 0.004T$.
$T = \frac{1}{0.004} = 250 \ K$.

(A) From the graph,we observe that the angle $\theta_1 < \theta_2$.
Since the slope of the $V-T$ graph is $\tan \theta = \frac{V}{T}$,it follows that $\tan \theta_1 < \tan \theta_2$.
Therefore,$\left( \frac{V}{T} \right)_1 < \left( \frac{V}{T} \right)_2$.
From the ideal gas equation $PV = \mu RT$,we have $\frac{V}{T} = \frac{\mu R}{P}$.
This implies that $\frac{V}{T} \propto \frac{1}{P}$.
Thus,$\frac{1}{P_1} < \frac{1}{P_2}$,which leads to $P_1 > P_2$.

(D) Let the initial state be $(P, V, T)$ and the final state be $(P, V', T')$.
Assuming the pressure remains constant (as it is a cylinder with a movable piston or open to atmosphere), we use the Ideal Gas Law: $PV = nRT$.
Initial state: $PV = n_1 RT$.
Final state: $PV' = n_2 R T'$.
Given: $T' = T + 0.20T = 1.2T$ and $V' = V - 0.10V = 0.9V$.
Substituting these into the final state equation: $P(0.9V) = n_2 R (1.2T)$.
$0.9 PV = n_2 R (1.2T) \Rightarrow n_2 = \frac{0.9}{1.2} n_1 = 0.75 n_1$.
The amount of gas leaked is $\Delta n = n_1 - n_2 = n_1 - 0.75 n_1 = 0.25 n_1$.
Percentage of gas leaked = $\frac{\Delta n}{n_1} \times 100 = 0.25 \times 100 = 25\%$.

(B) Given: Initial temperature $T_1 = 23^{\circ}C = 23 + 273 = 296 \, K$.
Initial volume $V_1 = 1500 \, ml$.
Final temperature $T_2 = 37^{\circ}C = 37 + 273 = 310 \, K$.
Since pressure and mass (amount of gas) are constant,we use Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Rearranging for $V_2$: $V_2 = V_1 \times \frac{T_2}{T_1}$.
Substituting the values: $V_2 = 1500 \times \frac{310}{296}$.
Calculating the result: $V_2 \approx 1570.9 \, ml$.

(B) The ideal gas equation is given by $PV = \mu RT$,where $\mu$ is the number of moles.
For $O_2$,the molar mass $M = 32 \, g/mol$.
The number of moles $\mu = \frac{\text{mass}}{\text{molar mass}} = \frac{8 \, g}{32 \, g/mol} = \frac{1}{4} \, mol$.
Substituting this into the ideal gas equation:
$PV = \left( \frac{1}{4} \right) RT = \frac{RT}{4}$.

(C) For a gas at constant volume,according to Gay-Lussac's Law,$P \propto T$.
This implies $\frac{\Delta P}{P} = \frac{\Delta T}{T}$.
Given that the percentage increase in pressure is $\frac{\Delta P}{P} \times 100 = 0.5\%$ and the change in temperature is $\Delta T = 5 \ K$.
Substituting these values into the equation:
$\frac{0.5}{100} = \frac{5}{T}$
$T = \frac{5 \times 100}{0.5}$
$T = 1000 \ K$.
Therefore,the initial temperature of the gas is $1000 \ K$.

(B) According to the ideal gas equation,$PV = \mu RT$,where $P$ is pressure,$V$ is volume,$\mu$ is the number of moles,$R$ is the universal gas constant,and $T$ is temperature.
Since the number of moles $\mu$,volume $V$,and temperature $T$ are the same for both gases,the pressure $P$ must also be the same.
Given that the pressure of $H_2$ is $4 \ atm$,the pressure of $He$ will also be $4 \ atm$.

(C) The ideal gas equation is given by $PV = NkT$,where $P$ is pressure,$V$ is volume,$N$ is the number of molecules,$k$ is the Boltzmann constant,and $T$ is the temperature in Kelvin.
Given: $P = 13.8 \ Pa$,$V = 1 \ m^3$,$k = 1.38 \times 10^{-23} \ JK^{-1}$,and $T = 27^{\circ}C = 27 + 273 = 300 \ K$.
Rearranging the formula for $N$: $N = \frac{PV}{kT}$.
Substituting the values: $N = \frac{13.8 \times 1}{1.38 \times 10^{-23} \times 300}$.
$N = \frac{13.8}{414 \times 10^{-23}} = \frac{13.8}{4.14 \times 10^{-21}} = \frac{13.8}{414} \times 10^{23} = 0.0333 \times 10^{23} = 3.33 \times 10^{21}$.
Thus,the number of molecules is approximately $3.3 \times 10^{21}$.

(A) Given mass of water $m = 4.5 \, kg = 4500 \, g$.
The molar mass of water $(H_2O)$ is $M = 18 \, g/mol$.
The number of moles $n = \frac{m}{M} = \frac{4500}{18} = 250 \, mol$.
At $STP$,the volume occupied by $1 \, mole$ of an ideal gas is $22.4 \, L$.
Therefore,the volume occupied by $250 \, moles$ is $V = n \times 22.4 \, L = 250 \times 22.4 = 5600 \, L$.
Since $1000 \, L = 1 \, m^3$,the volume is $V = 5.6 \, m^3$.

(D) Given: Initial pressure $P_1 = 1 \ atm$,Initial volume $V_1 = 100 \ ml$,Initial temperature $T_1 = 27 + 273 = 300 \ K$.
Final pressure $P_2 = 2 \ atm$,Final volume $V_2 = 100 \ ml$.
Since the volume remains constant $(V_1 = V_2)$,we use Gay-Lussac's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Substituting the values: $\frac{1}{300} = \frac{2}{T_2}$.
Solving for $T_2$: $T_2 = 300 \times 2 = 600 \ K$.
Converting to Celsius: $T_2(^{\circ}C) = 600 - 273 = 327^{\circ}C$.

(D) According to Charles's Law,at constant pressure,the volume $V$ of an ideal gas is directly proportional to its absolute temperature $T$,i.e.,$V = kT$,where $k$ is a constant.
Taking the derivative with respect to $T$,we get $\frac{dV}{dT} = k$.
The ratio of the increase in volume per unit rise in temperature to the original volume is given by $\frac{\frac{dV}{dT}}{V}$.
Substituting the values,we get $\frac{k}{kT} = \frac{1}{T}$.

(C) For a gas in a closed container,the volume $V$ is constant. According to Gay-Lussac's Law,$P \propto T$,where $P$ is pressure and $T$ is absolute temperature in Kelvin.
Taking the differential form,we have $\frac{\Delta P}{P} = \frac{\Delta T}{T}$.
Given,$\frac{\Delta P}{P} = 0.4 \% = \frac{0.4}{100}$ and $\Delta T = 1 \ K$ (since a change of $1 ^\circ C$ is equivalent to a change of $1 \ K$).
Substituting these values into the equation: $\frac{0.4}{100} = \frac{1}{T}$.
Solving for $T$: $T = \frac{100}{0.4} = 250 \ K$.

(C) The ideal gas equation is given by $PV = nRT$,where $P$ is pressure,$V$ is volume,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the absolute temperature.
For the first container: $P_1 V_1 = n_1 R T_1$. Given $n_1 = 1$,$T_1 = T$,and $P_1 = P$,we have $PV = RT$ (Equation $1$).
For the second container: $P_2 V_2 = n_2 R T_2$. Given $n_2 = 1$,$T_2 = 2T$,and the containers are identical,so $V_2 = V_1 = V$.
Substituting these values: $P_2 V = (1) R (2T) = 2RT$ (Equation $2$).
Dividing Equation $2$ by Equation $1$: $\frac{P_2 V}{PV} = \frac{2RT}{RT}$.
This simplifies to $\frac{P_2}{P} = 2$,which gives $P_2 = 2P$.

(B) According to Gay-Lussac's Law for a gas in a closed vessel (constant volume),$P \propto T$,which implies $\frac{\Delta P}{P} = \frac{\Delta T}{T}$.
Given: $\frac{\Delta P}{P} = 0.4 \ \% = 0.004$ and $\Delta T = 1 \ K$ (a change of $1 \ ^\circ C$ is equivalent to a change of $1 \ K$ in absolute temperature).
Substituting the values: $0.004 = \frac{1}{T}$.
Therefore,$T = \frac{1}{0.004} = 250 \ K$.
Converting to Celsius: $T( ^\circ C) = 250 - 273 = -23 \ ^\circ C$.
However,based on the standard interpretation of such problems where the result is expected in Kelvin or the temperature difference is treated as absolute,the initial temperature is $250 \ K$.

(C) According to Charles's Law,for an ideal gas at constant pressure,the volume $V$ is directly proportional to the absolute temperature $T$ $(V \propto T)$.
Given:
Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$.
Final temperature $T_2 = 327^{\circ}C = 327 + 273 = 600 \ K$.
Initial volume $V_1 = V$.
Using the relation $\frac{V_1}{T_1} = \frac{V_2}{T_2}$:
$\frac{V}{300} = \frac{V_2}{600}$.
$V_2 = V \times \frac{600}{300} = 2V$.
Therefore,the final volume is $2V$.

(C) The ideal gas equation is given by $PV = \mu RT$,where $\mu$ is the number of moles.
Number of moles $\mu = \frac{\text{mass}}{\text{molar mass}} = \frac{m}{M_0}$.
For oxygen gas $(O_2)$,the molar mass $M_0 = 32 \ g/mol$.
Given mass $m = 5 \ g$.
Substituting these values into the equation: $\mu = \frac{5}{32}$.
Therefore,the equation of state is $PV = \frac{5}{32} RT$.

(A) Given: Pressure $P = 1 \, atm = 1.01 \times 10^5 \, N \, m^{-2}$.
Volume $V = \text{Area} \times \text{Height} = 20 \, m^2 \times 3 \, m = 60 \, m^3$.
Temperature $T = 27 \, ^\circ C = 27 + 273 = 300 \, K$.
Universal gas constant $R = 8.31 \, J \, mol^{-1} K^{-1}$.
Molar mass of air $M = 29 \, g/mol = 0.029 \, kg/mol$.
Using the ideal gas equation $PV = nRT$,where $n$ is the number of moles:
$n = \frac{PV}{RT} = \frac{1.01 \times 10^5 \times 60}{8.31 \times 300} \approx 2430.8 \, mol$.
The mass of the air $m = n \times M = 2430.8 \times 0.029 \approx 70.49 \, kg \approx 70.5 \, kg$.

(C) The partial pressure of a gas in a mixture is given by $P_i = x_i P_{total}$,where $x_i$ is the mole fraction of the gas.
First,we calculate the mole fraction of $N_2$ from its mass percentage.
Let the total mass of air be $100 \,g$. The mass of $N_2$ is $75.5 \,g$ and the mass of the remaining air is $24.5 \,g$.
The average molar mass of air is approximately $29 \,g/mol$.
The number of moles of $N_2$ is $n_{N_2} = \frac{75.5}{28} \approx 2.696 \,mol$.
The total number of moles in $100 \,g$ of air is $n_{total} = \frac{100}{29} \approx 3.448 \,mol$.
The mole fraction of $N_2$ is $x_{N_2} = \frac{n_{N_2}}{n_{total}} = \frac{2.696}{3.448} \approx 0.782$.
Thus,the partial pressure of $N_2$ is $P_{N_2} = x_{N_2} \times P_{total} = 0.782 \times 1 \,atm \approx 0.78 \,atm$.

(B) From the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$.
So,$PV = \frac{m}{M} RT$,which gives $P = \frac{m}{V} \frac{RT}{M} = \rho \frac{RT}{M}$.
This implies $P \propto \rho T$,or $\frac{P_1}{P_2} = \frac{\rho_1 T_1}{\rho_2 T_2}$.
Given at point $A$: $P_A = P_0$,$T_A = T_0$,$\rho_A = \rho_0$.
Given at point $B$: $P_B = 3P_0$,$T_B = 2T_0$,$\rho_B = ?$.
Substituting the values: $\frac{P_0}{3P_0} = \frac{\rho_0 T_0}{\rho_B (2T_0)}$.
$\frac{1}{3} = \frac{\rho_0}{2 \rho_B} \Rightarrow \rho_B = \frac{3}{2} \rho_0$.

(D) Let the volume of each bulb be $V$. Initially,both bulbs are at $T_1 = 273 \ K$ and pressure $P$. The total number of moles $n$ is given by:
$n = \frac{PV}{RT_1} + \frac{PV}{RT_1} = \frac{2PV}{RT_1}$
When one bulb is at $T_1 = 273 \ K$ and the other is at $T_2$,the new pressure is $P' = 1.5P$. The total number of moles remains constant:
$n = \frac{P'V}{RT_1} + \frac{P'V}{RT_2} = \frac{1.5PV}{RT_1} + \frac{1.5PV}{RT_2}$
Equating the two expressions for $n$:
$\frac{2PV}{RT_1} = \frac{1.5PV}{RT_1} + \frac{1.5PV}{RT_2}$
$\frac{2}{273} = \frac{1.5}{273} + \frac{1.5}{T_2}$
$\frac{0.5}{273} = \frac{1.5}{T_2}$
$T_2 = 273 \times \frac{1.5}{0.5} = 273 \times 3 = 819 \ K$
Converting to Celsius: $T_2(^{\circ}C) = 819 - 273 = 546^{\circ}C$.

(D) The root mean square speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $M$ and $R$ are constant,$v_{rms} \propto \sqrt{T}$.
Given that the new root mean square speed $v'_{rms} = 2v_{rms}$,we have:
$\frac{v'_{rms}}{v_{rms}} = \sqrt{\frac{T'}{T}} \implies 2 = \sqrt{\frac{T'}{T}} \implies \frac{T'}{T} = 4 \implies T' = 4T$.
According to Charles's Law,for a fixed amount of gas at constant pressure,$V \propto T$.
Therefore,$\frac{V'}{V} = \frac{T'}{T} = 4$.
Thus,the new volume $V' = 4V$.

(B) Given:
Pressure $P = 2.5 \times 10^5 \, N \, m^{-2}$
Volume $V = 20 \, L = 20 \times 10^{-3} \, m^3$
Temperature $T = 27 \, ^\circ C = 27 + 273 = 300 \, K$
Molar mass $M = 32 \, g \, mol^{-1} = 32 \times 10^{-3} \, kg \, mol^{-1}$
Gas constant $R = 8.31 \, J \, mol^{-1} K^{-1}$
Using the ideal gas equation $PV = \mu RT$,where $\mu = \frac{m}{M}$:
$PV = \frac{m}{M} RT$
$m = \frac{PVM}{RT}$
Substituting the values:
$m = \frac{(2.5 \times 10^5) \times (20 \times 10^{-3}) \times (32 \times 10^{-3})}{8.31 \times 300}$
$m = \frac{2.5 \times 20 \times 32 \times 10^{-1}}{2493}$
$m = \frac{1600 \times 10^{-1}}{2493} \approx 0.06418 \, kg$
Rounding to the nearest significant value provided in the options,$m = 0.064 \, kg = 64 \, g$.

(A) Given:
Initial pressure $P_1 = 1 \,atm = 1.01 \times 10^5 \,N/m^2$.
Initial volume $V_1 = 1 \,m^3$.
Initial temperature $T_1 = 300 \,K$.
Final volume $V_2 = 2V_1 = 2 \,m^3$.
Final temperature $T_2 = 2T_1 = 600 \,K$.
Using the ideal gas law equation $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$:
$P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1}$
$P_2 = (1.01 \times 10^5) \times \frac{1}{2} \times \frac{600}{300}$
$P_2 = (1.01 \times 10^5) \times \frac{1}{2} \times 2$
$P_2 = 1.01 \times 10^5 \,N/m^2 \approx 10^5 \,N/m^2$.

(A) The initial pressure $P_1 = 15 \, atm = 15 \times 1.013 \times 10^5 \, Pa \approx 15.2 \times 10^5 \, Pa$. The initial temperature $T_1 = 27 + 273 = 300 \, K$. The volume $V = 30 \, L = 30 \times 10^{-3} \, m^3$.
Using the ideal gas law $PV = nRT = \frac{m}{M_w} RT$,the initial mass $m_1 = \frac{P_1 V M_w}{R T_1}$.
$m_1 = \frac{(15.2 \times 10^5) \times (30 \times 10^{-3}) \times (32 \times 10^{-3})}{8.31 \times 300} \approx 0.586 \, kg$.
After withdrawal,$P_2 = 11 \, atm = 11.14 \times 10^5 \, Pa$,$T_2 = 17 + 273 = 290 \, K$.
The final mass $m_2 = \frac{P_2 V M_w}{R T_2} = \frac{(11.14 \times 10^5) \times (30 \times 10^{-3}) \times (32 \times 10^{-3})}{8.31 \times 290} \approx 0.447 \, kg$.
The mass of oxygen taken out is $\Delta m = m_1 - m_2 = 0.586 - 0.447 = 0.139 \, kg \approx 0.14 \, kg$.

(D) Given: Initial mass $M_1 = 10 \, kg$,Initial pressure $P_1 = 10^7 \, N/m^2$,Final pressure $P_2 = 2.5 \times 10^6 \, N/m^2$.
Using the ideal gas equation $PV = nRT = \frac{M}{M_w} RT$,where $M$ is the mass of the gas.
Since $V$,$R$,$T$,and $M_w$ are constant,we have $P \propto M$.
Therefore,$\frac{P_2}{P_1} = \frac{M_2}{M_1}$.
Substituting the values: $\frac{2.5 \times 10^6}{10^7} = \frac{M_2}{10}$.
$0.25 = \frac{M_2}{10} \Rightarrow M_2 = 2.5 \, kg$.
The mass of gas removed is $\Delta M = M_1 - M_2 = 10 - 2.5 = 7.5 \, kg$.

(D) For an open container,the pressure $P$ and volume $V$ remain constant.
Using the ideal gas law $PV = nRT$,where $n$ is the number of moles,we have $n \propto 1/T$.
Therefore,$n_1 T_1 = n_2 T_2$.
Initial temperature $T_1 = 60 + 273 = 333 \ K$.
Let the initial number of moles be $n_1 = n$.
Since $1/4$ of the air escapes,the remaining moles $n_2 = n - (1/4)n = (3/4)n$.
Substituting the values: $n \times 333 = (3/4)n \times T_2$.
$T_2 = 333 \times (4/3) = 444 \ K$.
Converting back to Celsius: $T = 444 - 273 = 171^{\circ}C$.