Kinetic Theory of Gases MCQs with Answers – Part 1 (NEET Physics)

Explanation: The macroscopic form uses the amount of gas in moles, \(\mu\), together with the universal gas constant \(R\). Hence the macroscopic equation is \[ PV=\mu RT. \] The molecular form uses the actual number of molecules \(N\), so the appropriate constant is the Boltzmann constant \(k_B\): \[ PV=Nk_BT. \] The two forms are equivalent because \(N=\mu N_A\) and \(R=N_Ak_B\). Thus replacing \(N\) by \(\mu N_A\) in \(Nk_BT\) gives \(\mu RT\). Mixing \(N\) with \(R\), or \(\mu\) with \(k_B\), without the required Avogadro-number factor gives an inconsistent expression.

Explanation: The amount of gas and the vessel volume remain fixed. Therefore, from the ideal-gas equation, \[ \frac{P}{T}=\text{constant} \] Gas-law temperature ratios must use absolute temperature, not Celsius temperature. \[ T_1=27+273=300\,\text{K}, \] \[ T_2=127+273=400\,\text{K} \] Hence \[ \frac{P_2}{P_1}=\frac{T_2}{T_1}=\frac{400}{300}=\frac{4}{3} \] Using \(127/27\) would incorrectly treat Celsius readings as absolute temperatures. \(\textbf{Final answer:}\) \[ \boxed{\frac{P_2}{P_1}=\frac43} \]

Explanation: The number of molecules and number of moles are related by \[ N=\mu N_A \] Therefore, \[ \mu=\frac{N}{N_A} =\frac{1.204\times10^{24}}{6.02\times10^{23}} =2.0\,\text{mol} \] The macroscopic equation of state is \[ PV=\mu RT \] Substituting \(\mu=2.0\) gives \[ PV=2RT \] The Boltzmann constant \(k_B\) would be used with the molecule number \(N\), not directly with the mole number \(\mu\). \(\textbf{Final answer:}\) \[ \boxed{\mu=2.0\,\text{mol},\quad PV=2RT} \]

Explanation: For the same fixed amount of ideal gas, \[ \frac{PV}{T}=\text{constant} \] Thus \[ \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \] Substituting the given changes, \[ \frac{PV}{300} = \frac{\left(\frac32P\right)\left(\frac45V\right)}{T_2} \] The product of the pressure and volume factors is \[ \frac32\times\frac45=\frac65 \] Therefore, \[ \frac{T_2}{300}=\frac65, \] \[ T_2=360\,\text{K} \] The important point is that neither pressure nor volume is constant, so a single-variable proportionality cannot be applied independently. \(\textbf{Final answer:}\) \[ \boxed{T_2=360\,\text{K}} \]

Explanation: \(\textbf{First change:}\) At constant pressure for a fixed amount of ideal gas, \[ \frac{V}{T}=\text{constant} \] Hence \[ T'=300\left(\frac{1.2V}{V}\right)=360\,\text{K} \] \(\textbf{Second change:}\) The volume is now fixed at \(1.2V\), so \[ \frac{P}{T}=\text{constant} \] The pressure changes from \(P\) to \(0.75P\): \[ \frac{T_f}{360}=\frac{0.75P}{P}=0.75 \] Therefore, \[ T_f=(0.75)(360)=270\,\text{K} \] The two successive constraints must be applied separately; treating the whole change as either constant-pressure or constant-volume would give an incorrect result. \(\textbf{Final answer:}\) \[ \boxed{T_f=270\,\text{K}} \]

Explanation: For an ideal gas, the molecular form of the state equation is \[ P=nk_BT, \] where \(n=N/V\) is the number density. Therefore, \[ n=\frac{P}{k_BT} \] For the two states, \[ \frac{n_2}{n_1} = \frac{P_2}{P_1}\frac{T_1}{T_2} \] Substituting the given ratios, \[ \frac{n_2}{n_1} = \frac{1.20}{1.50} =0.80 \] Thus the final number density is \(80\%\) of the initial value, corresponding to a \(20\%\) decrease. The pressure increase alone does not determine the density change because the absolute temperature has also increased.

Explanation: For an ideal gas of molar mass \(M_m\), \[ P=\frac{\rho RT}{M_m} \] Hence \[ \rho=\frac{PM_m}{RT} \] For gas X, \[ \rho_X=\frac{P(2M)}{RT} =\frac{2PM}{RT} \] For gas Y, \[ \rho_Y= \frac{\left(\frac32P\right)M}{R(3T)} = \frac{PM}{2RT} \] Therefore, \[ \frac{\rho_X}{\rho_Y} = \frac{2PM/(RT)}{PM/(2RT)} =4 \] The density comparison depends simultaneously on pressure, absolute temperature and molar mass; comparing molar masses alone would miss the altered state conditions. \(\textbf{Final answer:}\) \[ \boxed{\frac{\rho_X}{\rho_Y}=4} \]

Explanation: Statement I is correct because \(PV=\mu RT\), and \(\mu\) is constant for a fixed amount of gas. Thus \(PV/T=\mu R\) remains constant for that sample. Statement II is also correct because the molecular equation can be written as \[ n=\frac{P}{k_BT}. \] At specified \(P\) and \(T\), this expression is independent of the chemical identity of the ideal gas. Statement III is incorrect because at fixed temperature \[ P=nk_BT, \] so pressure is directly, not inversely, proportional to number density. Doubling \(n\) at fixed \(T\) doubles \(P\). Statement IV is correct: at lower pressure and higher temperature, molecules are generally farther apart and intermolecular interaction effects become less significant. Therefore, Statements I, II and IV are the correct set.

Explanation: For an ideal gas, \[ PV=Nk_BT \] Therefore, the molecule number is proportional to \[ N\propto\frac{PV}{T} \] Let \(N_0\) denote the number of molecules in state A. For any observed state, \[ \frac{N}{N_0} = \frac{(P/P_0)(V/V_0)}{T/T_0} \] At state A, the ratio is \(1\). For state B, \[ \frac{N_B}{N_0} = \frac{1.5\times0.8}{1.2} =1 \] For state C, \[ \frac{N_C}{N_0} = \frac{0.8\times1.5}{1.2} =1 \] Thus A, B and C contain the same number of molecules despite their different pressures, volumes and temperatures. For state D, \[ \frac{N_D}{N_0} = \frac{1.2\times1.0}{1.6} =0.75 \] Hence the reduction in molecule number first appears between C and D. The fraction of initial molecules lost is \[ 1-0.75=0.25 \] Therefore, \(25\%\) of the initial molecules were lost. \(\textbf{Final answer:}\) Leakage occurred between states C and D, with \(25\%\) of the initial molecules lost.

Explanation: \(\textbf{Compare the mole numbers:}\) For each vessel, \[ \mu=\frac{PV}{RT} \] For X, \[ \mu_X=\frac{P(2V)}{R(300)} \] For Y, \[ \mu_Y=\frac{(2P)V}{R(600)} \] Hence \[ \frac{\mu_X}{\mu_Y} = \frac{2/300}{2/600} =2 \] Therefore, \[ \mu_X:\mu_Y=2:1 \] The total amount is \(5.0\,\text{mol}\), so \[ \mu_X=\frac{2}{3}(5.0)=\frac{10}{3}\,\text{mol} \] \(\textbf{Convert moles to molecules:}\) \[ N_X=\mu_XN_A =\frac{10}{3}\left(6.0\times10^{23}\right) \] Thus \[ N_X=2.0\times10^{24} \] The comparison must first account for all three state variables before the mole-to-molecule conversion is made. \(\textbf{Final answer:}\) \[ \boxed{N_X=2.0\times10^{24}} \]

Explanation: \(\textbf{Amount initially in vessel A:}\) \[ \mu_A=\frac{PV}{R(300)} \] \(\textbf{Amount initially in vessel B:}\) \[ \mu_B=\frac{(2P)(2V)}{R(600)} =\frac{4PV}{600R} =\frac{2PV}{300R} \] Hence the total amount is \[ \mu_{\text{tot}} = \frac{PV}{300R}+\frac{2PV}{300R} = \frac{3PV}{300R} = \frac{PV}{100R} \] No gas is lost when the vessels are connected. The total rigid volume is \[ V_{\text{tot}}=V+2V=3V \] At the final temperature \(400\,\text{K}\), \[ P_f(3V)=\mu_{\text{tot}}R(400) \] Substituting \(\mu_{\text{tot}}=PV/(100R)\), \[ P_f(3V)=\frac{PV}{100R}R(400)=4PV \] Therefore, \[ P_f=\frac{4P}{3} \] The initial pressures cannot simply be averaged because the two vessels initially contain different amounts of gas. \(\textbf{Final answer:}\) \[ \boxed{P_f=\frac{4P}{3}} \]

Explanation: \(\textbf{Initial molecule number:}\) Using \(PV=Nk_BT\), \[ N_i=\frac{P_0V}{k_B(300)} \] \(\textbf{Molecules remaining after heating and leakage:}\) The vessel still has volume \(V\), while its final pressure is \(P_0\) and temperature is \(450\,\text{K}\): \[ N_f=\frac{P_0V}{k_B(450)} \] Thus \[ \frac{N_f}{N_i}=\frac{300}{450}=\frac23 \] Therefore the fraction that leaked out is \[ \frac{N_{\ell}}{N_i}=1-\frac23=\frac13 \] Hence \[ N_{\ell}=\frac13\frac{P_0V}{k_B(300)} \] \(\textbf{State of the collected gas:}\) If its collected volume is \(V_c\), \[ \left(\frac{P_0}{2}\right)V_c=N_{\ell}k_B(300) \] Substituting \(N_{\ell}\), \[ \left(\frac{P_0}{2}\right)V_c = \frac13P_0V \] Therefore, \[ V_c=\frac{2V}{3} \] The calculation requires first finding how much gas leaked and then applying the state equation again to the collected portion. \(\textbf{Final answer:}\) \[ \boxed{V_c=\frac{2V}{3}} \]

Explanation: \(\textbf{Find the molecular number density:}\) For an ideal gas, \[ P=nk_BT \] Therefore, \[ n=\frac{P}{k_BT} \] Substituting, \[ n= \frac{1.38\times10^5} {(1.38\times10^{-23})(300)} \] Thus \[ n=3.33\times10^{25}\,\text{m}^{-3} \] \(\textbf{Find the mass of one molecule:}\) Mass density is the mass of all molecules per unit volume: \[ \rho=nm, \] where \(m\) is the mass of one molecule. Hence \[ m=\frac{\rho}{n} = \frac{1.00}{3.33\times10^{25}} \approx3.0\times10^{-26}\,\text{kg} \] \(\textbf{Convert molecular mass to molar mass:}\) \[ M=mN_A \] Therefore, \[ M=(3.0\times10^{-26})(6.0\times10^{23}) =1.8\times10^{-2}\,\text{kg mol}^{-1} \] Since \(1.8\times10^{-2}\,\text{kg mol}^{-1}=18\,\text{g mol}^{-1}\), the required molar mass is \(18\,\text{g mol}^{-1}\). \(\textbf{Final answer:}\) \[ \boxed{M=18\,\text{g mol}^{-1}} \]

Explanation: \(\textbf{Compare the temperatures:}\) The density form of the ideal-gas equation is \[ P=\frac{\rho RT}{M_m} \] Rearranging, \[ T=\frac{PM_m}{\rho R} \] The two gases have the same \(P\) and \(\rho\), so their temperatures are directly proportional to molar mass: \[ \frac{T_Y}{T_X}=\frac{4M}{M}=4 \] \(\textbf{Compare molecular number densities:}\) For an ideal gas, \[ n=\frac{P}{k_BT} \] Since the pressures are equal, \[ \frac{n_Y}{n_X}=\frac{T_X}{T_Y} \] Therefore, \[ \frac{n_Y}{n_X}=\frac14 \] Thus the heavier gas must be at four times the absolute temperature to have the same pressure and mass density, and consequently it has one-fourth the molecular number density. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{T_Y}{T_X},\frac{n_Y}{n_X}\right)=\left(4,\frac14\right)} \]

Explanation: \(\textbf{Interpret the graph slope:}\) For a sealed ideal gas in a rigid vessel, \[ P=\frac{Nk_B}{V}T \] Therefore the slope of a \(P\)-versus-\(T\) graph is \[ s=\frac{Nk_B}{V} \] Hence \[ \frac{s_X}{s_Y} = \frac{N_X/V_X}{N_Y/V_Y} = \frac{N_X}{N_Y}\frac{V_Y}{V_X} \] Given \[ \frac{s_X}{s_Y}=\frac32,\qquad \frac{V_X}{V_Y}=\frac43, \] we obtain \[ \frac{N_X}{N_Y} = \frac32\frac43 =2 \] \(\textbf{Compare mass densities:}\) Because X and Y contain the same gas, the molecular mass is the same, so total gas mass is proportional to molecule number. Thus \[ \frac{\rho_X}{\rho_Y} = \frac{N_X/V_X}{N_Y/V_Y} \] This is also exactly the slope ratio: \[ \frac{\rho_X}{\rho_Y}=\frac32 \] Therefore the required ordered pair is \(\textbf{Final answer:}\) \[ \boxed{\left(2,\frac32\right)} \]

Explanation: \(\textbf{Convert the density change into a volume change:}\) For a fixed mass \(m_g\), \[ \rho=\frac{m_g}{V} \] The final density is \[ \rho_2=0.80\rho_1 \] Therefore, \[ \frac{V_2}{V_1} = \frac{\rho_1}{\rho_2} = \frac{1}{0.80} = \frac54 \] \(\textbf{Use the ideal-gas state equation:}\) The amount of gas is fixed, so \[ \frac{PV}{T}=\text{constant} \] Hence \[ \frac{T_2}{T_1} = \frac{P_2V_2}{P_1V_1} \] The pressure increases by \(50\%\), giving \[ \frac{P_2}{P_1}=1.50=\frac32 \] Thus \[ \frac{T_2}{T_1} = \frac32\times\frac54 = \frac{15}{8} \] The gas simultaneously has a higher pressure and a larger volume, so its absolute temperature must rise by a factor greater than either isolated density or pressure change would suggest. Therefore, \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{T_2}{T_1},\frac{V_2}{V_1}\right) = \left(\frac{15}{8},\frac54\right)} \]

Explanation: \(\textbf{Extract the graph slope:}\) For a sealed ideal-gas sample at fixed temperature, \[ P=\frac{Nk_BT}{V}=(Nk_BT)\left(\frac1V\right) \] Hence the slope of the \(P\)-versus-\(1/V\) graph is \[ s=Nk_BT \] Therefore, \[ \frac{s_X}{s_Y}=\frac{N_XT_X}{N_YT_Y} \] Using \(s_X/s_Y=3\) and \(T_X/T_Y=3/2\), \[ 3=\frac{N_X}{N_Y}\left(\frac32\right), \] so \[ \frac{N_X}{N_Y}=2 \] \(\textbf{Convert total number to number density:}\) Since \(n=N/V\), \[ \frac{n_X}{n_Y} =\frac{N_X}{N_Y}\frac{V_Y}{V_X} =2\left(\frac12\right)=1 \] Thus the samples contain different total numbers of molecules but have equal number densities at the selected states. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{N_X}{N_Y},\frac{n_X}{n_Y}\right)=(2,1)} \]

Explanation: The Assertion is correct because an ideal gas is a limiting model that real gases approximate under suitable conditions. Lower pressure generally corresponds to a more dilute gas, so the average separation between molecules becomes large compared with molecular dimensions. With molecules farther apart, the effects associated with intermolecular forces become less important for most of the molecular motion. Raising the temperature also makes the kinetic effects of molecular motion increasingly dominant relative to intermolecular interaction effects. These changes make the assumptions underlying the perfect-gas equation more nearly valid. The Reason therefore states the physical basis for the improved ideal-gas approximation. The Reason is not merely another independent fact; it directly explains why lowering pressure and raising temperature favour ideal behaviour. This does not mean that an ordinary real gas becomes mathematically perfect under all such conditions; rather, its behaviour approaches the ideal-gas model more closely. Hence both statements are true and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Use equal pressure first:}\) For an ideal gas, \[ P=nk_BT \] Since \(P_X=P_Y\), \[ n_XT_X=n_YT_Y \] Given \(n_X=2n_Y\), \[ 2n_YT_X=n_YT_Y \] Therefore, \[ \frac{T_X}{T_Y}=\frac12 \] \(\textbf{Connect number density with mass density:}\) The mass of one molecule is \(M/N_A\), so \[ \rho=n\frac{M}{N_A} \] Hence \[ \frac{\rho_X}{\rho_Y} = \frac{n_X}{n_Y}\frac{M_X}{M_Y} \] Substituting the given density ratios, \[ 3=2\frac{M_X}{M_Y}, \] which gives \[ \frac{M_X}{M_Y}=\frac32 \] \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{T_X}{T_Y},\frac{M_X}{M_Y}\right) = \left(\frac12,\frac32\right)} \]

Explanation: \(\textbf{Effect of leakage on number density:}\) The vessel is rigid, so its volume remains constant. After \(20\%\) of the molecules escape, \[ N_2=0.80N_1 \] Since \(n=N/V\) and \(V\) is unchanged, \[ \frac{n_2}{n_1}=\frac{N_2}{N_1}=0.80 \] \(\textbf{Use the molecular ideal-gas equation:}\) For a fixed vessel volume, \[ P=nk_BT \] Therefore, \[ \frac{P_2}{P_1} = \frac{n_2T_2}{n_1T_1} \] Substituting \(P_2/P_1=1.20\) and \(n_2/n_1=0.80\), \[ 1.20=0.80\frac{T_2}{T_1} \] Hence \[ \frac{T_2}{T_1}=\frac{1.20}{0.80}=1.50 \] The pressure can increase despite loss of molecules because the temperature rise more than compensates for the reduced number density. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{n_2}{n_1},\frac{T_2}{T_1}\right)=(0.80,1.50)} \]