A gas mixture consists of molecules
Question: A gas mixture consists of molecules of tyres A, B, and C with masses mA mB mC. Rank the three types of molecules in decreasing order of (a) average KE (b) rms speeds Solution: (a) Pressure and temperature are the same, therefore, KEc KEb KEa (b) When P and T are constant, (Vrms)c (Vrms)b (Vrms)a...
Read More →The container shown in the figure has two chambers,
Question: The container shown in the figure has two chambers, separated by a partition, of volumes V1= 2.0 litre and V2= 3.0 litre. The chambers contain 1= 4.0 and 2= 5.0 moles of a gas at pressure p1= 1.00 atm and p2= 2.00 atm. Calculate the pressure after the partition is removed and the mixture attains equilibrium. Solution: For an ideal gas, PV = RT P1V1 = 1R1T1 P2V2 = 2R2T2 P1 = 1 atm P2 = 2 atm V1 = 2L V2 = 3L T1 = T = T2 1 = 4 2 = 5 Substituting all the values we get, P = 1.6 atm...
Read More →If A + B + C = π, prove that
Question: If A + B + C = , prove that $\cos 2 A-\cos 2 B+\cos 2 C=1-4 \sin A \cos B \sin C$ Solution: $=\cos 2 A-\cos 2 B+\cos 2 C$ Using, $\cos A-\cos B=2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{B-A}{2}\right)$ $=\cos 2 A-\left\{2 \sin \left(\frac{2 B+2 C}{2}\right) \sin \left(\frac{2 B-2 C}{2}\right)\right\}$ $=\cos 2 A-\{2 \sin (B+C) \sin (B-C)\}$ since $A+B+C=\pi$ $\rightarrow \mathrm{B}+\mathrm{C}=180-\mathrm{A}$ And $\sin (\pi-A)=\sin A$ $=\cos 2 A-\{2 \sin (\pi-A) \sin (B-C)\}$ $...
Read More →Calculate the ratio of the mean free paths
Question: Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters 1 A and 2 A. The gases may be considered under identical conditions of temperature, pressure, and volume. Solution: The ratio of the mean free paths of the molecules is 4:1....
Read More →A gas mixture consists of 2.0 moles
Question: A gas mixture consists of 2.0 moles of oxygen and 4.0 moles of neon a temperature T. Neglecting all vibrational modes, calculate the total internal energy of the system. (Oxygen has two rotational modes.) Solution: Degree of freedom of oxygen = 5 Total internal energy of 2-mole oxygen = 5RT Total internal energy of 4 mole of Ne = 6RT The total internal energy of the system = 11 RT...
Read More →Two molecules of a gas have speeds
Question: Two molecules of a gas have speeds of 9 106m/s and 1 106m/s respectively. What is the root mean square speed of these molecules? Solution: V1 = 9 106m/s V2 = 1 106m/s Vrms = 6.4 106m/s...
Read More →Find the The Slopes of the tangent and the normal to the following curves at the indicated points:
Question: Find the The Slopes of the tangent and the normal to the following curves at the indicated points: $x=a(\theta-\sin \theta), y=a(1+\cos \theta)$ at $\theta=-\pi / 2$ Solution: Given: $x=a(\theta-\sin \theta) \ y=a(1+\cos \theta)$ at $\theta=\frac{-\pi}{2}$ Here, To find $\frac{d y}{d x}$, we have to find $\frac{d y}{d \theta} \ \frac{d x}{d \theta}$ and and divide $\frac{\frac{d y}{d x}}{\frac{d x}{d \theta}}$ and we get our desired $\frac{d y}{d x}$. $\therefore \frac{\mathrm{dy}}{\ma...
Read More →The molecules of a given mass of a gas
Question: The molecules of a given mass of a gas have root mean square speeds of 100 m/s at 27oC and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at 127oC and 2.0 atmospheric pressure? Solution: Vrms = 100 m/s T1 = 300 K T2 = 400 K Vrms = 3RT/M V2rms =? Vrms = 115.4 m/s...
Read More →The volume of a given mass
Question: The volume of a given mass of a gas at 27oC, 1 atm is 100 cc. What will be its volume at 327oC? Solution: T1 = 27oC T1 = 300 K V1 = 100 cm3 We know that V is proportional to T V/T = constant V1/T1 = V2/T2 V2 = V1(T2/T1) V2 = 200 cc...
Read More →Calculate the number of atoms
Question: Calculate the number of atoms in 39.4 g gold. Molar mass of gold is 197 g/mole. Solution: Molar mass = 6.023 1023atoms Mass of gold, m = 39.4 g Molar mass of gold, M = 197 g/mol 196 g of gold contains 6.023 1023atoms 39.4 g of gold contains 1.20 1023atoms....
Read More →If A + B + C = π, prove that
Question: If A + B + C = , prove that $\cos 2 A-\cos 2 B-\cos 2 C=-1+4 \cos A \sin B \sin C$ Solution: $=\cos 2 A-(\cos 2 B+\cos 2 C)$ Using formula $\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$ $=\cos 2 A-\left\{2 \cos \left(\frac{2 B+2 C}{2}\right) \cos \left(\frac{2 B-2 C}{2}\right)\right\}$ $=\cos 2 A-\{2 \cos (B+C) \cos (B-C)\}$ Since $A+B+C=\pi$ $\rightarrow \mathrm{B}+\mathrm{C}=180-\mathrm{A}$ $=\cos 2 \mathrm{~A}-\{2 \cos (\pi-\mathrm{A}) \cos (\mathr...
Read More →When an ideal gas is compressed adiabatically,
Question: When an ideal gas is compressed adiabatically, its temperature rises: the molecules on the average have more kinetic energy than before. The kinetic energy increases, (a) because of collisions with moving parts of the wall only (b) because of collisions with the entire wall (c) because the molecules gets accelerated in their motion inside the volume (d) because of the redistribution of energy amongst the molecules Solution: The correct answer is (a) because of collisions with moving pa...
Read More →Ina diatomic molecules,
Question: Ina diatomic molecules, the rotational energy at a given temperature (a) obeys Maxwells distribution (b) have the same value for all molecules (c) equals the translational kinetic energy for each molecule (d) is 2/3rdthe translational kinetic energy for each molecule Solution: The correct answer is (a) obeys Maxwells distribution (d) is 2/3rdthe translational kinetic energy for each molecule...
Read More →Diatomic molecules like hydrogen have
Question: Diatomic molecules like hydrogen have energies due to both translational as well as rotational motion. From the equation in kinetic theory pV = 2/3 E, E is (a) the total energy per unit volume (d) only the translational part of energy because rotational energy is very small compared to the translational energy (c) only the translational part of the energy because during collisions with the wall pressure relates to change in linear momentum (d) the translational part of the energy becau...
Read More →Find the The Slopes of the tangent and the normal to the following curves at the indicated points :
Question: Find the The Slopes of the tangent and the normal to the following curves at the indicated points : $y=2 x^{2}+3 \sin x$ at $x=0$ Solution: Given: $y=2 x^{2}+3 \sin x$ at $x=0$ First, we have to find $\frac{d y}{d x}$ of given function, $f(x)$, i.e, to find the derivative of $f(x)$ $\therefore \frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \cdot \mathrm{x}^{\mathrm{n}-1}$ The Slope of the tangent is $\frac{d y}{d x}$ $\Rightarrow \mathrm{y}=2 \mathrm{x}^...
Read More →An inflated rubber balloon contains
Question: An inflated rubber balloon contains one mole of an ideal gas, has a pressure p, volume V, and temperature T. If the temperature rises to 1.1T and the volume is increased to 1.05V, the final pressure will be (a) 1.1 p (b) p (c) less than p (d) between p and 1.1 Solution: The correct answer is (d) between p and 1.1...
Read More →A vessel of volume V contains a mixture of 1 mole
Question: A vessel of volume V contains a mixture of 1 mole of hydrogen and 1 mole of oxygen. Let f1(v)dv, denote the fraction of molecules with speed between v and (v + dv) with f2(v)dv similarly for oxygen. Then (a) f1(v) + f2(v) = f(v) obeys the Maxwells distribution law (b) f1(v), f2(v) will obey the Maxwells distribution law separately (c) neither f1(v) nor f2(v) will obey the Maxwells distribution law (d) f2(v) and f1(v) will be the same Solution: The correct answer is b) f1(v), f2(v) will...
Read More →If A + B + C = π, prove that
Question: If A + B + C = , prove that $\sin 2 A+\sin 2 B-\sin 2 C=4 \cos A \cos B \sin C$ Solution: $=\sin 2 A+\sin 2 B-\sin 2 C$ $=2 \sin (B+C) \cos A+2 \sin (A+C) \cos B-2 \sin (A+B) \cos C$ Using formula, $\sin (A+B)=\sin A \cos B+\cos A \sin B$ $=\sin 2 \mathrm{~A}+\sin 2 \mathrm{~B}-\sin 2 \mathrm{C}$ Using formula sin2A = 2sinAcosA = 2sinAcosA + 2sinBcosB - 2sinCcosC Since A + B + C = $\rightarrow \mathrm{B}+\mathrm{C}=180-\mathrm{A}$ And $\sin (\pi-A)=\sin A$ $=2 \sin (B+C) \cos A+2 \sin ...
Read More →1 mole of H2 gas is contained in
Question: 1 mole of H2 gas is contained in a box of volume V = 1.00 m3 at T = 300 K. The gas is heated to a temperature of T = 3000 K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (a) same as the pressure initially (b) 2 times the pressure initially (c) 10 times the pressure initially (d) 20 times the pressure initially Solution: The correct answer is (d) 20 times the pressure initially...
Read More →Find the The Slopes of the tangent and the normal to the following curves at the indicated points:
Question: Find the The Slopes of the tangent and the normal to the following curves at the indicated points: $y=x^{3}-x$ at $x=2$ Solution: Given: $y=x^{3}-x$ at $x=2$ First, we have to find $\frac{d y}{d x}$ of given function, $f(x)$, i.e, to find the derivative of $f(x)$ $\therefore \frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \cdot \mathrm{x}^{\mathrm{n}-1}$ The Slope of the tangent is $\frac{d y}{d x}$ $\Rightarrow y=x^{3}-x$ $\Rightarrow \frac{d y}{d x}=\fr...
Read More →Boyle’s law is applicable for an
Question: Boyles law is applicable for an (a) adiabatic process (b) isothermal process (c) isobaric process (d) isochoric process Solution: The correct answer is (b) isothermal process...
Read More →A cubic vessel (with face horizontal + vertical)
Question: A cubic vessel (with face horizontal + vertical) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of500 ms1inverticaldirection. The pressure of the gas inside the vessel as observed by us on the ground (a) remains the same because 500 ms1is very much smaller thanvrmsof the gas. (b) remains the same becausemotionof the vessel as a whole does not affect the relative motion of the gas molecules and the walls. (c) will increase by a factor eq...
Read More →Find the The Slopes of the tangent and the normal to the following curves at the indicated points:
Question: Find the The Slopes of the tangent and the normal to the following curves at the indicated points: $y=\sqrt{x}$ at $x=9$ Solution: Given: $y=\sqrt{x}$ at $x=9$ First, we have to find $\frac{d y}{d x}$ of given function, $f(x)$,i.e, to find the derivative of $f(x)$ $\Rightarrow y=\sqrt{x}$ $\therefore \sqrt[n]{x}=x^{\frac{1}{n}}$ $\Rightarrow y=(x)^{\frac{1}{2}}$ $\therefore \frac{d y}{d x}\left(x^{n}\right)=n \cdot x^{n-1}$ The Slope of the tangent is $\frac{d y}{d x}$ $\Rightarrow y=(...
Read More →Consider that an ideal gas is expanding
Question: Consider that an ideal gas is expanding in a process given by P = f(V), which passes through a point (V0, P0). Show that the gas is absorbing heat at (P0, V0) if the slope of the curve P = f(V) is larger than the slope of the adiabat passing through (P0, V0). Solution: Slope of the graph, $\left(V_{0}, P_{0}\right)=\left(\frac{\mathrm{d} P}{\mathrm{~d} V}\right)_{V_{0}, P_{0}}$ Using the above relation, we can find that $V_{0} f^{\prime}\left(V_{0}\right)-\gamma P_{0}$ $f^{\prime}\left...
Read More →If the coefficient of performance
Question: If the coefficient of performance of a refrigerator is 5 and operates at the room temperature, find the temperature inside the refrigerator. Solution: To find the temperature inside the refrigerator, we need to find the work done W = 0 Q1= Q2 = Where is the coefficient of performance Given, T1= 27 + 273 = 300 K Coefficient of performance of refrigerator = 5 Coefficient of performance is given as $\beta=\frac{Q_{2}}{Q_{1}-Q_{2}}$ Using the above equation, the temperature can be determin...
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