A polyatomic molecule has 3 translational, 3 rotational degrees of freedom. The molar specific heat ratio \((\gamma = \frac{C_p}{C_v})\) for this gas is
Total degrees of freedom \(f = 3 + 3 = 6\). Thus, \(C_v = \frac{f}{2} R = 3R\), and \(C_p = C_v + R = 4R\). Therefore, \(\gamma = \frac{C_p}{C_v} = \frac{4}{3}\).
The efficiency of heat engine working between \(27^\circ\text{C}\) and \(327^\circ\text{C}\) may be
The maximum possible efficiency (Carnot efficiency) is \(eta_{\text{max}} = 1 - \frac{T_L}{T_H} = 1 - \frac{300}{600} = 50\%\). Practical heat engines have efficiency less than or equal to 50%, so 50%, 35%, and 25% are all possible values.
A body cools from \(80^\circ\text{C}\) to \(50^\circ\text{C}\) in 6 minutes. The time it takes to cool from \(60^\circ\text{C}\) to \(40^\circ\text{C}\) is (The temperature of surrounding is \(20^\circ\text{C}\))
Using Newton's law of cooling: \(\frac{T_1 - T_2}{t} = K \left[ \frac{T_1 + T_2}{2} - T_0 \right]\). For the first case, \(\frac{30}{6} = K[65 - 20] ⇒ 5 = 45K ⇒ K = \frac{1}{9}\). For the second case, \(\frac{20}{t} = \frac{1}{9}[50 - 20] = \frac{30}{9} = \frac{10}{3} ⇒ t = 6 \text{ minutes}\).
The ratio of \(v_{\text{rms}} : v_{\text{mp}} : v_{\text{avg}}\) is (symbols have their usual meaning)
\[v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\], \[v_{\text{mp}} = \sqrt{\frac{2RT}{M}}\], and \[v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}}\]. Therefore, \[v_{\text{rms}} : v_{\text{mp}} : v_{\text{avg}} = \sqrt{3} : \sqrt{2} : \sqrt{\frac{8}{\pi}}\].
The work done by 2 moles of polyatomic gas \( \gamma = \frac{4}{3} \) initially at room temperature to increase its volume eight time during adiabatic process will be (Take \(R = 2\text{ cal mol}^{-1}\text{ K}^{-1}\) and room temperature 27°C)
In adiabatic process, \(T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}\). Here, \(\gamma - 1 = 1/3\). Since \(V_2/V_1 = 8\), we get \(T_2 = T_1 (1/8)^{1/3} = T_1/2 = 300/2 = 150\text{ K}\). The work done is \(W = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{2 \times 2 \times (300 - 150)}{1/3} = 1800\text{ cal}\).
The percentage change in length of 1 m iron rod if its temperature changes by 100ºC is (\(\alpha\) for iron is \(2 \times 10^{-5}/\text{ºC}\))
The percentage change in length is given by \(\frac{\Delta L}{L} \times 100 = \alpha \Delta T \times 100 = (2 \times 10^{-5}) \times 100 \times 100 = 0.2%\).
A gas mixture consists of 5 moles of oxygen and 3 moles of argon at temperature T. Assuming the gases to be ideal and oxygen bond to be rigid, the total internal energy of the mixture is (where R denotes the universal gas constant)
Internal energy is \( U = n \frac{f}{2} RT \). For diatomic \( \text{O}_2 \) (rigid, \( f_1=5 \)), \( U_1 = 5 \times \frac{5}{2} RT = 12.5 RT \). For monoatomic \( \text{Ar} \) (\( f_2=3 \)), \( U_2 = 3 \times \frac{3}{2} RT = 4.5 RT \). Total \( U = 12.5RT + 4.5RT = 17RT \).
Consider the following statements:
(A) Temperature of a body is related to its average internal energy, not to the kinetic energy of motion of its centre of mass.
(B) The first law of thermodynamics is based on law of conservation of energy applied to any system in which energy transfer from or to the surrounding (through heat and work) is taken into account.
Based on above information, pick correct option.
Statement (A) is correct because temperature measures internal, disordered molecular kinetic energy, not ordered bulk kinetic energy. Statement (B) is also correct because the first law of thermodynamics is simply the law of conservation of energy.
Consider the following statements:
(A) Convection involves flow of matter within a fluid due to unequal temperatures of its parts.
(B) The change of solid state to vapour state without passing through liquid state is called sublimation.
Based on above information, pick correct option.
Statement A is correct as thermal convection is driven by density variations arising from unequal temperature distributions in fluids. Statement B is also correct as sublimation is the direct phase transition from solid to gas.
Thin copper wire of length \( L \) increases in length by 2% when heated from \( T_1 \) to \( T_2 \). If a copper cube having side \( 10L \) is heated from \( T_1 \) to \( T_2 \) then the percentage change in volume of the cube is
The percentage change in length is \( \frac{\Delta L}{L} \times 100 = 2% \). Since volume expansion coefficient is three times the linear expansion coefficient (\( \gamma = 3\alpha \)), the percentage change in volume is \( 3 \times 2% = 6% \).