Ideal gas equation and Related gas laws Questions in English

(A) Boyle's law states that for a fixed amount of an ideal gas at constant temperature,$PV = k$ (where $k$ is a constant).
Taking the derivative of both sides with respect to the variables:
$d(PV) = d(k)$
Using the product rule: $P dV + V dP = 0$
Rearranging the terms: $V dP = -P dV$
Dividing both sides by $PV$: $\frac{dP}{P} = - \frac{dV}{V}$.

(D) The ideal gas equation is given by $PV = nRT$.
Since the number of moles $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we can write $PV = \frac{m}{M}RT$.
Rearranging the terms to solve for density $(d = \frac{m}{V})$,we get $d = \frac{m}{V} = \frac{PM}{RT}$.
Thus,the correct option is $D$.

(B) The density $(d)$ of an ideal gas is given by the formula $d = \frac{PM}{RT}$,where $P$ is pressure,$M$ is molar mass,$R$ is the gas constant,and $T$ is temperature.
Since $M$ and $R$ are constant,$d \propto \frac{P}{T}$.
Calculating the ratio $\frac{P}{T}$ for each option:
$A$: $\frac{0.5}{600} \approx 0.00083$
$B$: $\frac{2}{150} \approx 0.0133$
$C$: $\frac{1}{300} \approx 0.0033$
$D$: $\frac{1}{500} = 0.002$
The ratio $\frac{P}{T}$ is maximum for option $B$.

(B) The density $(d)$ of an ideal gas is given by the formula $d = \frac{PM}{RT}$,where $P$ is pressure,$M$ is molar mass,$R$ is the gas constant,and $T$ is temperature in Kelvin.
Since $M$ and $R$ are constants,$d \propto \frac{P}{T}$.
For option $A$ $(S.T.P.)$: $P = 1\,atm$,$T = 273\,K$,so $d \propto \frac{1}{273} \approx 0.0036$.
For option $B$: $P = 2\,atm$,$T = 273\,K$,so $d \propto \frac{2}{273} \approx 0.0073$.
For option $C$: $P = 1\,atm$,$T = 546\,K$,so $d \propto \frac{1}{546} \approx 0.0018$.
For option $D$: $P = 2\,atm$,$T = 546\,K$,so $d \propto \frac{2}{546} \approx 0.0036$.
Comparing the values,the density is highest at ${0\,^oC}$ and $2\,atm$.

(C) According to the ideal gas equation,$PV = nRT = (\frac{m}{M})RT$.
Rearranging for density $(d = \frac{m}{V})$,we get $d = \frac{PM}{RT}$.
Since $P$,$R$,and $T$ are constant,$d \propto M$.
Therefore,$\frac{d_A}{d_B} = \frac{M_A}{M_B}$.
Given $d_A = 3 \, d_B$ and $M_A = M$,we have $\frac{3 \, d_B}{d_B} = \frac{M}{M_B}$.
Solving for $M_B$,we get $M_B = \frac{M}{3}$.

(B) The root mean square velocity $(U_{rms})$ is given by the formula $U_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $U_{rms} \propto \sqrt{T}$,we have $\frac{U_1}{U_2} = \sqrt{\frac{T_1}{T_2}}$.
Given $U_1 = 5 \times 10^4 \ cm \ s^{-1}$ and $U_2 = 10 \times 10^4 \ cm \ s^{-1}$,then $\frac{U_1}{U_2} = \frac{1}{2}$.
Squaring both sides,$\frac{T_1}{T_2} = (\frac{1}{2})^2 = \frac{1}{4}$,which means $T_2 = 4T_1$.
For an ideal gas in a fixed volume,$P \propto T$ (Gay-Lussac's Law). Since the temperature is quadrupled,the pressure also becomes four times the initial value.

(C) The compressibility factor $Z$ is defined as $Z = \frac{PV}{RT}$.
For an ideal gas,the ideal gas equation is $PV = nRT$. For $1 \text{ mole}$ of gas,$PV = RT$.
Substituting this into the expression for $Z$,we get $Z = \frac{RT}{RT} = 1$.
Therefore,the compressibility factor of an ideal gas is $1$.

(B) The compressibility factor $Z$ is defined as $Z = \frac{PV}{nRT}$.
For an ideal gas,the equation of state is $PV = nRT$.
Substituting this into the expression for $Z$,we get $Z = \frac{nRT}{nRT} = 1$.
Therefore,the compressibility factor for an ideal gas is $1$.

(B) In an ideal gas,there are no intermolecular forces of attraction between the particles.
During unrestrained expansion (expansion against vacuum),no work is done by the gas $(w = 0)$.
Since there are no attractive forces to overcome,no internal energy is consumed to separate the molecules,and therefore,the temperature of the ideal gas remains constant.

(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.
At this temperature,the compressibility factor $Z$ remains close to $1$ for a wide range of pressure.

(C) According to Avogadro's Law,at the same temperature and pressure,equal volumes of all gases contain an equal number of molecules.
Given that $1 \, L$ of $O_2$ contains $N$ molecules at a specific temperature and pressure.
Therefore,$1 \, L$ of $SO_2$ will also contain $N$ molecules under the same conditions.
Consequently,$2 \, L$ of $SO_2$ will contain $2 \times N = 2N$ molecules.

(B) Using the ideal gas equation,$PV = nRT$.
Given: $P = 760 \ mm \ Hg = 1 \ atm$,$V = 145 \ L$,$T = 27 \ ^oC = 300 \ K$,and $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.
Calculating the number of moles $(n)$: $n = \frac{PV}{RT} = \frac{1 \times 145}{0.082 \times 300} \approx 5.89 \ mol \approx 6 \ mol$.
The molar mass of $H_2$ is $2 \ g/mol$.
Therefore,the mass of $H_2 = n \times \text{molar mass} = 6 \ mol \times 2 \ g/mol = 12 \ g$.

(A) Using the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$.
For butane $(C_4H_{10})$,the molar mass $M = (4 \times 12) + (10 \times 1) = 58\, g/mol$.
Given $m = 1\, g$,$T = 350\, K$,$P = 1\, atm$,and $R = 0.0821\, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.
$V = \frac{nRT}{P} = \frac{m}{M} \times \frac{RT}{P} = \frac{1}{58} \times \frac{0.0821 \times 350}{1} \approx 0.4956\, L$.
Since $1\, L = 1000\, cm^3$,the volume is $0.4956 \times 1000 \approx 495.6\, cm^3$.
The closest option is $495\, cm^3$.

(B) According to Charles's Law,at constant pressure,$\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = V$,$T_1 = 27 + 273 = 300\,K$,$V_2 = 2V$.
Substituting the values: $\frac{V}{300} = \frac{2V}{T_2}$.
$T_2 = 300 \times 2 = 600\,K$.
Converting to Celsius scale: $T_2 = 600 - 273 = 327\,^oC$.

(A) Number of moles of $O_2 = \frac{4 \ g}{32 \ g/mol} = 0.125 \ mol$.
Number of moles of $H_2 = \frac{2 \ g}{2 \ g/mol} = 1 \ mol$.
Total number of moles $(n)$ = $0.125 + 1 = 1.125 \ mol$.
Using the ideal gas equation $PV = nRT$,where $R = 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1}$ and $T = 273 \ K$:
$P = \frac{nRT}{V} = \frac{1.125 \times 0.0821 \times 273}{1} \approx 25.215 \ atm$.

(D) According to Boyle's Law,at constant temperature,$P_1V_1 = P_2V_2$.
Given: $V_1 = 2 \ L$,$P_2 = 2P_1$.
Substituting the values: $P_1 \times 2 \ L = (2P_1) \times V_2$.
$V_2 = \frac{P_1 \times 2 \ L}{2P_1} = 1 \ L$.
Therefore,the volume becomes $1 \ L$.

(C) According to Charles's Law,at constant pressure,$V \propto T$,so $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = 300 \; cm^{3}$,$T_1 = 27 + 273 = 300 \; K$,$T_2 = 37 + 273 = 310 \; K$.
Calculating $V_2$: $V_2 = V_1 \times \frac{T_2}{T_1} = 300 \; cm^{3} \times \frac{310 \; K}{300 \; K} = 310 \; cm^{3}$.
The volume of air expelled is $\Delta V = V_2 - V_1 = 310 \; cm^{3} - 300 \; cm^{3} = 10 \; cm^{3}$.

(C) According to Charles's Law,at constant pressure,$V \propto T$ or $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = 300 \ mL$,$T_1 = 27 + 273 = 300 \ K$,$T_2 = -3 + 273 = 270 \ K$.
Substituting the values: $V_2 = \frac{T_2}{T_1} \times V_1 = \frac{270 \ K}{300 \ K} \times 300 \ mL = 270 \ mL$.

(A) The internal energy $(U)$ of an ideal gas is a function of temperature only,i.e.,$U = f(T)$.
Since the process occurs at constant temperature $(T)$,the change in internal energy $(\Delta U)$ is zero.
Therefore,the internal energy remains the same.

(A) The universal gas constant $R$ has different values depending on the units used.
For pressure in $atm$ and volume in $L$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
In $SI$ units,where pressure is in $Pa$ $(N/m^2)$ and volume is in $m^3$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Comparing the given options,option $A$ represents the correct value in $L \ atm \ K^{-1} \ mol^{-1}$.

(B) For an ideal gas,the internal energy $(U)$ is a function of temperature $(T)$ only.
It does not depend on the volume $(V)$ of the gas.
Therefore,the change in internal energy with respect to volume at constant temperature is zero,i.e.,${\left( {\frac{{dU}}{{dV}}} \right)_T = 0}$.

(C) According to Avogadro's Law,equal volumes of all gases at the same temperature and pressure contain an equal number of molecules.
Since the volume,temperature,and pressure are the same for both $O_2$ and $H_2$,the number of molecules will be equal.

(B) Using the ideal gas equation: $PV = nRT$
Given:
$P = 760 \ mm \ Hg = 1 \ atm$
$V = 500 \ cm^3 = 0.5 \ L$
$T = 300 \ K$
$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$
Substituting the values:
$1 \ atm \times 0.5 \ L = n \times 0.0821 \ L \ atm \ K^{-1} \ mol^{-1} \times 300 \ K$
$n = \frac{0.5}{0.0821 \times 300}$
$n = \frac{0.5}{24.63} \approx 0.0203 \ mol$
$n = 2.03 \times 10^{-2} \ mol$

(B) From the ideal gas equation,$PV = nRT$.
Since $n = \frac{m}{M_w}$ (where $m$ is mass and $M_w$ is molecular weight),we can write $PV = \frac{m}{M_w} RT$.
Rearranging the equation,we get $P M_w = \frac{m}{V} RT$.
Since density $d = \frac{m}{V}$,the equation becomes $P M_w = dRT$.
Therefore,$d = \frac{P M_w}{RT}$.

(A) The ideal gas equation is $PV = nRT$.
Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M}RT$.
Rearranging for pressure,$P = \frac{m}{V} \cdot \frac{RT}{M}$.
Since density $d = \frac{m}{V}$,the equation becomes $P = d \cdot \frac{RT}{M}$.
Comparing this with the equation of a straight line $y = mx$,where $y = P$ and $x = d$,the slope is $m = \frac{RT}{M}$.
Since $R$ and $M$ are constants,the slope is directly proportional to temperature $T$.
From the graph,the slope of the line for $T_1$ is greater than the slope of the line for $T_2$.
Therefore,$T_1 > T_2$.

(D) Given: $wt = 5 \, g$,$V = 1 \, L$,$P = 10 \, atm$,Molar mass of ethane $(C_2H_6)$ $M_w = 30 \, g/mol$.
Using the ideal gas equation: $PV = nRT = \frac{wt}{M_w} RT$.
Substituting the values: $10 \times 1 = \frac{5}{30} \times 0.0821 \times T$.
$T = \frac{10 \times 30}{5 \times 0.0821} = \frac{300}{0.4105} \approx 730.81 \, K$.
Converting to Celsius: $T(^oC) = T(K) - 273.15 = 730.81 - 273.15 = 457.66 \, ^oC$.
Rounding to the nearest provided option,the correct value is $457.81 \, ^oC$.

(C) Given: $T_1 = 27 + 273 = 300 \ K$,$V_1 = 0.418 \ L$,$P_1 = \frac{740}{760} \ atm$.
At $STP$: $T_2 = 273 \ K$,$P_2 = 1 \ atm$,$V_2 = ?$.
Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Substituting the values: $\frac{(\frac{740}{760}) \times 0.418}{300} = \frac{1 \times V_2}{273}$.
$V_2 = \frac{740 \times 0.418 \times 273}{760 \times 300} \approx 0.37 \ L$.

(A) The universal gas constant $R$ has different values depending on the units used for pressure,volume,and temperature.
For the $SI$ unit system,where pressure is in $Pa$ $(N \ m^{-2})$,volume is in $m^3$,and temperature is in $K$,the value of $R$ is $8.314 \ J \ K^{-1} \ mol^{-1}$.
Since $1 \ J = 1 \ Pa \ m^3$,the unit $J \ K^{-1} \ mol^{-1}$ is the correct $SI$ unit for $R = 8.314$.

(B) Given:
Mass of gas $(w)$ = $120 \ g$
Molar mass $(M)$ = $40 \ g \ mol^{-1}$
Volume $(V)$ = $20 \ L$
Temperature $(T)$ = $400 \ K$
Gas constant $(R)$ = $0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$
Step $1$: Calculate the number of moles $(n)$:
$n = \frac{w}{M} = \frac{120 \ g}{40 \ g \ mol^{-1}} = 3 \ mol$
Step $2$: Use the ideal gas equation $PV = nRT$ to find pressure $(P)$:
$P = \frac{nRT}{V}$
$P = \frac{3 \ mol \times 0.0821 \ L \ atm \ K^{-1} \ mol^{-1} \times 400 \ K}{20 \ L}$
$P = \frac{3 \times 0.0821 \times 400}{20}$
$P = 3 \times 0.0821 \times 20$
$P = 60 \times 0.0821 = 4.926 \ atm$
Rounding to the nearest provided option,the pressure is $4.92 \ atm$.

(A) Given: $w = 7 \, g$,$P = \frac{750}{760} \, atm$,$M_w = 28 \, g/mol$,$T = 27 + 273 = 300 \, K$.
Using the ideal gas equation: $PV = nRT = \frac{w}{M_w} RT$.
Substituting the values: $\frac{750}{760} \times V = \frac{7}{28} \times 0.0821 \times 300$.
$V = \frac{0.25 \times 0.0821 \times 300 \times 760}{750} \approx 6.29 \, L$.
Thus,the volume is approximately $6.3 \, L$.

(B) The density $(d)$ of a gas is given by the ideal gas equation: $d = \frac{PM_w}{RT}$
Given:
Pressure $(P)$ = $\frac{800}{760} \, atm$
Molar mass $(M_w)$ of $CO_2$ = $44 \, g/mol$
Temperature $(T)$ = $100 + 273 = 373 \, K$
Gas constant $(R)$ = $0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$
Substituting the values:
$d = \frac{(\frac{800}{760}) \times 44}{0.0821 \times 373}$
$d \approx 1.5124 \, g/L$

(A) Initially,both parts $A$ and $B$ have equal volumes $(V)$ and equal pressures $(P = 1 \ atm)$.
When the partition is removed,the total volume becomes $2V$.
According to Boyle's Law,$P_1V_1 = P_2V_2$.
For the total system,the final pressure $P_f$ is calculated as: $P_f = \frac{P_A V + P_B V}{2V} = \frac{1 \times V + 1 \times V}{2V} = 1 \ atm$.
Since the gases mix and occupy the total volume,the partial pressure of each gas decreases to $0.5 \ atm$,but the total pressure of the mixture remains $1 \ atm$ throughout the container.

(C) At $S.T.P.$ (Standard Temperature and Pressure),the molar volume of an ideal gas is $22.4 \ L \ mol^{-1}$.
The molar mass of nitrogen monoxide $(NO)$ is $14 + 16 = 30 \ g \ mol^{-1}$.
The density $(d)$ is calculated using the formula: $d = \frac{\text{Molar Mass}}{\text{Molar Volume}}$.
$d = \frac{30 \ g \ mol^{-1}}{22.4 \ L \ mol^{-1}} \approx 1.34 \ g \ L^{-1}$.

(A) According to the ideal gas law,$PV = nRT$,which implies $P = \frac{nRT}{V}$.
Thus,the pressure $P$ is directly proportional to temperature $T$ $(P \propto T)$.
Therefore,the statement that pressure is independent of temperature is incorrect.
Option $B$ is correct because average kinetic energy of gas molecules is $KE = \frac{3}{2}RT$,which depends only on temperature.
Option $C$ is correct because real gases deviate from ideal behavior at high pressure and low temperature.
Option $D$ is correct because the Boltzmann constant $k_B = \frac{R}{N_A}$ represents the gas constant per molecule.

(C) The ideal gas equation is $PV = nRT$,where $n = \frac{m}{M}$.
Substituting $n$,we get $PV = \frac{m}{M} RT$.
Since density $d = \frac{m}{V}$,we can write $P = \frac{dRT}{M}$.
For gas $A$: $P_A = \frac{d_A RT}{M_A}$.
For gas $B$: $P_B = \frac{d_B RT}{M_B}$.
The ratio is $\frac{P_A}{P_B} = \frac{d_A}{d_B} \times \frac{M_B}{M_A}$.
Given $d_A = 3.0 \, g \, dm^{-3}$,$d_B = 1.5 \, g \, dm^{-3}$,and $M_A = \frac{1}{2} M_B$ (or $\frac{M_B}{M_A} = 2$).
Substituting these values: $\frac{P_A}{P_B} = \frac{3.0}{1.5} \times 2 = 2 \times 2 = 4$.

(D) The universal gas constant $R$ has different values depending on the units used:
$1$. In $SI$ units,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$.
$2$. In $CGS$ units,$1 \, J = 10^7 \, \text{ergs}$,so $R = 8.314 \times 10^7 \, \text{ergs} \, K^{-1} \, mol^{-1}$.
$3$. Since $1 \, J = 1 \, N \cdot m$ and $1 \, N = 10^5 \, \text{dynes}$,$1 \, J = 10^5 \, \text{dynes} \times 10^2 \, cm = 10^7 \, \text{dynes} \cdot cm$. Thus,$8.314 \times 10^7 \, \text{dynes} \cdot cm \, K^{-1} \, mol^{-1}$ is also correct.
$4$. In $L \cdot atm$ units,$R = 0.0821 \, L \cdot atm \, K^{-1} \, mol^{-1}$.
Therefore,the value $8.31 \, L \cdot atm \, K^{-1} \, mol^{-1}$ is incorrect.

(B) Using Boyle's Law,$P_1V_1 = P_2V_2$ at constant temperature.
For gas $A$: $1000 \ torr \times 500 \ mL = P_A \times 2000 \ mL$. So,$P_A = (1000 \times 500) / 2000 = 250 \ torr$.
For gas $B$: $800 \ torr \times 1000 \ mL = P_B \times 2000 \ mL$. So,$P_B = (800 \times 1000) / 2000 = 400 \ torr$.
According to Dalton's Law of partial pressures,the total pressure $P_{total} = P_A + P_B$.
$P_{total} = 250 \ torr + 400 \ torr = 650 \ torr$.

(D) The ideal gas equation is given by $PV = nRT$,where $R$ is the universal gas constant.
$R$ is a universal constant,meaning its value does not depend on the nature,pressure,or temperature of the gas.
However,the numerical value of $R$ depends on the units used for pressure $(P)$,volume $(V)$,and temperature $(T)$.
For example,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$ when using liters and atmospheres,but $R = 8.314 \ J \ K^{-1} \ mol^{-1}$ in $SI$ units.

(D) Using the ideal gas equation in terms of density: $d = \frac{PM_w}{RT}$
Given: $P = \frac{780}{760} \ atm$,$T = -23 + 273 = 250 \ K$,$d = 1.40 \ g/L$,$R = 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1}$
Substituting the values: $1.40 = \frac{(\frac{780}{760}) \times M_w}{0.0821 \times 250}$
$M_w = \frac{1.40 \times 0.0821 \times 250 \times 760}{780} \approx 28 \ g/mol$
The molar mass of $N_2$ is $28 \ g/mol$. Therefore,the gas is $N_2$.

(B) According to the ideal gas law,$PV = nRT$. Since the amount of gas $(n)$ is constant,the relationship is $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$.
At $STP$,$P_1 = 1 \ atm$,$T_1 = 273 \ K$,and $V_1 = 10 \ L$.
We want $V_2 = 10 \ L$,so the condition $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ must be satisfied.
For option $A$: $\frac{1}{273} \neq \frac{2}{273}$.
For option $B$: $T_2 = 273 + 273 = 546 \ K$. $\frac{P_2}{T_2} = \frac{2}{546} = \frac{1}{273}$. This matches the initial condition.
Therefore,option $B$ is correct.

(C) Given: For the unknown gas,$w_1 = 3.7 \ g$,$T_1 = 25 + 273 = 298 \ K$.
For $H_2$ gas,$w_2 = 0.184 \ g$,$T_2 = 17 + 273 = 290 \ K$,$M_2 = 2 \ g/mol$.
Since pressure and volume are the same,we use the ideal gas law $PV = nRT = (w/M)RT$.
Thus,$(w_1/M_1)T_1 = (w_2/M_2)T_2$.
Substituting the values: $(3.7/M_1) \times 298 = (0.184/2) \times 290$.
$M_1 = (3.7 \times 298 \times 2) / (0.184 \times 290) = 2205.2 / 53.36 \approx 41.326 \ g/mol$.

(B) Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
Given:
$P_1 = \frac{730}{760} \ atm$,$V_1 = 600 \ mL = 0.6 \ L$,$T_1 = 27 + 273 = 300 \ K$
At $STP$:
$P_2 = 1 \ atm$,$T_2 = 273 \ K$
Substituting the values:
$\frac{(\frac{730}{760}) \times 0.6}{300} = \frac{1 \times V_2}{273}$
$V_2 = \frac{730 \times 0.6 \times 273}{760 \times 300} \approx 0.524 \ L$

(B) According to the ideal gas equation,$PV = nRT = (m/M)RT$,where $P$ is pressure,$V$ is volume,$m$ is mass,$M$ is molar mass,$R$ is the gas constant,and $T$ is temperature.
Rearranging for density $(d = m/V)$,we get $P = (dRT)/M$.
Since $R$,$T$,and $M$ are constant for a given gas at a constant temperature,the pressure $P$ is directly proportional to the density $d$ $(P \propto d)$.
Given that the initial density $d_1 = 1.458 \, mg/L$ at $P_1 = 1 \, atm$,if the new density $d_2 = 2 \times d_1$,then the new pressure $P_2$ will be $2 \times P_1$.
Therefore,$P_2 = 2 \times 1 \, atm = 2 \, atm$.

(C) According to Charles's Law,at constant pressure,$V \propto T$ or $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = 300 \ mL$,$T_1 = 27 \ ^oC = 27 + 273 = 300 \ K$,$T_2 = -3 \ ^oC = -3 + 273 = 270 \ K$.
Substituting the values: $\frac{300}{300} = \frac{V_2}{270}$.
Therefore,$V_2 = 270 \ mL$.

(A) Using the ideal gas equation: $PV = nRT = \frac{w}{M_w} RT$
Rearranging for volume: $V = \frac{wRT}{P M_w}$
Given values: $w = 1 \ g$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,$T = 350 \ K$,$P = 1 \ atm$,and molar mass of butane $(C_4H_{10}) = 4 \times 12 + 10 \times 1 = 58 \ g/mol$.
Substituting the values: $V = \frac{1 \times 0.0821 \times 350}{1 \times 58} \approx 0.495 \ L$
Since $1 \ L = 1000 \ cm^3$,$V = 0.495 \times 1000 = 495 \ cm^3$.

(A) According to Charles's Law,at constant pressure,$V \propto T$,which means $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given that the volume increases by $10\%$,$V_2 = V_1 + 0.10V_1 = 1.10V_1$.
Substituting this into the equation: $\frac{V_1}{T_1} = \frac{1.10V_1}{T_2}$.
Solving for $T_2$: $T_2 = 1.10T_1$.
The percentage increase in temperature is $\frac{T_2 - T_1}{T_1} \times 100 = \frac{1.10T_1 - T_1}{T_1} \times 100 = 0.10 \times 100 = 10\%$.