Ideal gas equation and Related gas laws Questions in English

(A) For a fixed mass $(m)$ of an ideal gas,the ideal gas equation is $pV = nRT$,where $n$ is constant.
$(A)$ For $p$ vs $1/V$ (Boyle's Law): $p = (nRT) \times (1/V)$. The slope is $nRT$. Since $T_1 > T_2 > T_3$,the slope $T_1 > T_2 > T_3$. This graph is correct.
$(B)$ For $p$ vs $T$ (Gay-Lussac's Law): $p = (nR/V) \times T$. The slope is $nR/V$. Since $V_1 > V_2 > V_3$,the slope $1/V_1 < 1/V_2 < 1/V_3$. Thus,the order of slopes should be $V_3 > V_2 > V_1$. The provided graph shows $V_1 > V_2 > V_3$,which is incorrect.
$(C)$ For $V$ vs $p$ (Boyle's Law): $V = (nRT) \times (1/p)$. This is a rectangular hyperbola,not a straight line passing through the origin. The provided graph is incorrect.
Therefore,the correct representation is $(A)$.

(C) According to Charles's Law, at constant pressure, $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = 2 \,L$, $T_1 = 273 \,K$ (at $STP$), $V_2 = 4 \,L$.
Substituting the values: $\frac{2}{273} = \frac{4}{T_2}$.
$T_2 = \frac{273 \times 4}{2} = 546 \,K$.
Converting to Celsius: $546 \,K - 273 = 273^{\circ} C$.

(B) Given: $W = 4 \ g$,$V = 5.6035 \ L$,$T = 546 \ K$,$P = 2 \ atm$.
Using the ideal gas equation: $PV = nRT$,where $n = \frac{W}{M}$.
$PV = \frac{W}{M} RT$
$M = \frac{WRT}{PV}$
Substituting the values: $M = \frac{4 \times 0.0821 \times 546}{2 \times 5.6035}$
$M = \frac{179.3304}{11.207} \approx 16 \ g/mol$.

(D) Initial state for gas $A$: $P_1 = 8 \ atm$,$V_1 = 12 \ L$. Total volume after opening the stop cock: $V_{total} = 12 \ L + 8 \ L = 20 \ L$.
Since the temperature is constant,we use Boyle's Law $(P_1V_1 = P_2V_2)$ for each gas independently.
For gas $A$: $8 \ atm \times 12 \ L = P_A \times 20 \ L \implies P_A = \frac{96}{20} = 4.8 \ atm$.
For gas $B$: $5 \ atm \times 8 \ L = P_B \times 20 \ L \implies P_B = \frac{40}{20} = 2.0 \ atm$.
Thus,the partial pressures of $A$ and $B$ are $4.8 \ atm$ and $2.0 \ atm$ respectively.

(C) The partial pressure of a gas in a mixture is the pressure it would exert if it alone occupied the entire volume of the container at the same temperature.
Using the ideal gas equation,$PV = nRT$,the partial pressure of $N_2$ $(p_{N_2})$ is given by $p_{N_2} = \frac{n_{N_2} R T}{V}$.
Given:
$n_{N_2} = 2.5 \ mol$
$V = 24.6 \ L$
$T = 300 \ K$
$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$
Substituting the values:
$p_{N_2} = \frac{2.5 \ mol \times 0.0821 \ L \ atm \ K^{-1} \ mol^{-1} \times 300 \ K}{24.6 \ L}$
$p_{N_2} = \frac{61.575}{24.6} \ atm$
$p_{N_2} = 2.5 \ atm$.

(D) The $RMS$ velocity of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{T}$,reducing the $RMS$ velocity to half means the final temperature $T_f$ must be $\frac{1}{4}$ of the initial temperature $T_i$ $(T_f = \frac{T_i}{4})$.
For a reversible adiabatic process,$T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$.
For a diatomic gas,$\gamma = \frac{7}{5}$,so $\gamma - 1 = \frac{2}{5}$.
Substituting the values: $T_i V_i^{2/5} = (\frac{T_i}{4}) V_f^{2/5}$.
This simplifies to $4 = (\frac{V_f}{V_i})^{2/5}$.
Raising both sides to the power of $\frac{5}{2}$: $4^{5/2} = \frac{V_f}{V_i}$.
$V_f / V_i = (2^2)^{5/2} = 2^5 = 32$.

(B) Using the ideal gas law for a fixed amount of gas: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
Given: $P_1 = 203 \ kPa$,$V_1 = 0.5 \ m^3$,$T_1 = 400 \ K$
$P_2 = 304 \ kPa$,$V_2 = 0.2 \ m^3$
Substituting the values: $\frac{203 \times 0.5}{400} = \frac{304 \times 0.2}{T_2}$
$T_2 = \frac{304 \times 0.2 \times 400}{203 \times 0.5} = \frac{24320}{101.5} \approx 239.6 \ K$
Change in temperature $\Delta T = |T_2 - T_1| = |239.6 - 400| = 160.4 \ K$
The nearest integer is $160 \ K$.

(C) The formula for reversible isothermal work is: $W = -2.303 nRT \log \left(\frac{V_2}{V_1}\right)$
Given: $n = 2.5 \ mol$,$V_1 = 2 \ dm^3$,$V_2 = 20 \ dm^3$,$W = -16.5 \ kJ = -16500 \ J$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$
Substituting the values: $-16500 = -2.303 \times 2.5 \times 8.314 \times T \times \log \left(\frac{20}{2}\right)$
$-16500 = -2.303 \times 2.5 \times 8.314 \times T \times \log(10)$
Since $\log(10) = 1$,we have: $-16500 = -2.303 \times 2.5 \times 8.314 \times T$
$-16500 = -47.87 \times T$
$T = \frac{16500}{47.87} \approx 344.68 \ K$
Rounding to the nearest value,$T \approx 345 \ K$.

(C) For an adiabatic process,the relation between pressure and volume is $Pv^{\gamma} = K$.
Differentiating both sides,we get $v^{\gamma} dP + P \gamma v^{\gamma-1} dv = 0$.
Rearranging the terms,we get $\frac{dP}{P} = -\gamma \frac{dv}{v}$.
To find the percentage change,multiply both sides by $100$:
$\frac{dP}{P} \times 100 = -\gamma \left( \frac{dv}{v} \times 100 \right)$.
Given $\gamma = 1.4$ and $\frac{dv}{v} \times 100 = 5 \%$,we substitute these values:
$\text{Percentage change in pressure} = -1.4 \times 5 = -7 \%$.
The negative sign indicates a decrease in pressure. Thus,the magnitude of the percentage change is $7 \%$.

(D) From the graph,at point $x$: $T_1 = 300 \ K, V_1 = 20 \ L$.
At point $z$: $T_2 = 600 \ K, V_2 = 40 \ L$.
According to the ideal gas law,$pV = nRT$,which implies $\frac{pV}{T} = nR$.
Since $n$ and $R$ are constants,$\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}$.
Substituting the values: $\frac{p_1 \times 20}{300} = \frac{p_2 \times 40}{600}$.
$\frac{p_1}{15} = \frac{p_2}{15}$,which gives $p_1 = p_2$.
Since the pressure remains constant during the process from $z$ to $x$,it is an isobaric process.

(B) Avogadro's law states that equal volumes of all gases at the same temperature and pressure contain an equal number of molecules. This law is strictly applicable only to $ideal$ gases,as it assumes that gas molecules have negligible volume and no intermolecular forces of attraction,which are the fundamental postulates of the kinetic molecular theory of gases.

(D) Mass of gas initially $= 30.0 \ kg - 15.0 \ kg = 15.0 \ kg$.
Mass of gas remaining $= 24.2 \ kg - 15.0 \ kg = 9.2 \ kg$.
Mass of gas used $= 15.0 \ kg - 9.2 \ kg = 5.8 \ kg = 5800 \ g$.
Molar mass of butane $(C_4H_{10})$ $= 4 \times 12 + 10 \times 1 = 58 \ g/mol$.
Number of moles of gas used $(n)$ $= \frac{5800 \ g}{58 \ g/mol} = 100 \ mol$.
Using the ideal gas equation $PV = nRT$ at $1 \ atm$ and $27^{\circ}C$ $(300 \ K)$:
$V = \frac{nRT}{P} = \frac{100 \ mol \times 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} \times 300 \ K}{1 \ atm} = 2463 \ L$.
Since $1 \ m^3 = 1000 \ L$,the volume in cubic metres is $\frac{2463}{1000} \ m^3 = 2.463 \ m^3$.

(D) The ideal gas equation is given by $PV = nRT = \frac{w}{M} RT$.
Rearranging for density $(d = \frac{w}{V})$,we get $d = \frac{PM}{RT}$.
Thus,the molar mass $M = \frac{dRT}{P}$.
Given: $d = 1.964 \ g \ L^{-1}$,$T = 273 \ K$,$P = 76 \ cm \ Hg = 1 \ atm$,and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
Substituting the values: $M = \frac{1.964 \times 0.0821 \times 273}{1} \approx 44 \ g \ mol^{-1}$.
The molar mass of $CO_2$ is $12 + 2 \times 16 = 44 \ g \ mol^{-1}$.
Therefore,the gas is $CO_2$.

(A) For an ideal gas,the ideal gas equation is $pV = \frac{w}{M}RT$,where $w$ is the mass,$M$ is the molecular weight,$R$ is the gas constant,and $T$ is the temperature.
Since $w$,$R$,and $T$ are constant,let $wRT = K$. Then $pV = \frac{K}{M}$,or $V = \frac{K}{Mp}$.
Taking the logarithm on both sides: $\log V = \log \left( \frac{K}{M} \right) - \log p$.
This is in the form of $y = mx + c$,where $y = \log V$,$x = \log p$,$m = -1$,and the intercept $c = \log \left( \frac{K}{M} \right)$.
From the graph,the intercept for $M_{2}$ is higher than the intercept for $M_{1}$.
Therefore,$\log \left( \frac{K}{M_{2}} \right) > \log \left( \frac{K}{M_{1}} \right)$.
This implies $\frac{K}{M_{2}} > \frac{K}{M_{1}}$,which means $M_{1} > M_{2}$.

(A) For a reversible adiabatic process,the relation between pressure $(P)$ and temperature $(T)$ is given by $P^{1-\gamma} T^\gamma = \text{constant}$.
This can be rearranged as $P \propto T^{\frac{\gamma}{\gamma-1}}$.
Given: $P \propto T^3$.
Therefore,$\frac{\gamma}{\gamma-1} = 3$.
$\gamma = 3\gamma - 3$.
$2\gamma = 3$.
$\gamma = \frac{3}{2}$.
Since $\gamma = \frac{C_p}{C_V}$,the ratio is $\frac{3}{2}$.

(C) The ideal gas equation is $pV = nRT$.
For $n = 1 \ mol$,the equation becomes $V = (\frac{R}{p})T$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = V$ and $x = T$,the slope $m$ is given by $\frac{R}{p}$.
Given that the slope is $X$ at a pressure $p = 2 \ atm$,we have $X = \frac{R}{2}$.
Therefore,$R = 2X \ L \ atm \ mol^{-1} \ K^{-1}$.