A vessel contains gas A at pressure \(P\), volume \(V\) and temperature \(T\). Another vessel contains gas B at pressure \(2P\), volume \(\frac{V}{2}\) and temperature \(2T\). Ratio of number of molecules of B to A will be
1. \(\frac{1}{2}\)
2. 2
3. 4
4. \(\frac{1}{4}\)
View Answer
From the ideal gas law, \(PV = N k_B T\) which gives \(N = \frac{PV}{k_B T}\). Taking the ratio: \(\frac{N_B}{N_A} = \frac{P_B V_B / T_B}{P_A V_A / T_A} = \frac{2P times (V/2) / 2T}{PV / T} = \frac{1}{2}\).
Consider statements (A) and (B) given below:
A. Thermodynamics deals with the process of conversion of heat into work only.
B. Heat given to a system and work done by the system are state variables in thermodynamics.
Choose the correct option.
1. Statement (A) is correct but statement (B) is incorrect
2. Statement (A) is incorrect but statement (B) is correct
3. Both statements are correct
4. Both statements are incorrect
View Answer
Statement A is incorrect because thermodynamics also deals with other energy conversions. Statement B is incorrect because heat and work are path functions, not state variables.
The sun delivers about \( 1400 \text{ W/m}^2 \) of electromagnetic flux to Earth’s surface. If the flux falls on a roof of dimensions \( 8\text{ m} \times 20\text{ m} \) normally, then total power incident on the roof will be
1. 224 kW
2. 112 kW
3. 32 kW
4. 96 kW
View Answer
Power is given by the product of intensity and area: \( P = I \times A \). Here \( I = 1400 \text{ W/m}^2 \) and \( A = 8\text{ m} \times 20\text{ m} = 160 \text{ m}^2 \). Therefore, \( P = 1400 \times 160 = 224,000 \text{ W} = 224 \text{ kW} \).
When 150 gm of ice at \( 0^\circ\text{C} \) is mixed with 60 gm of water at \( 50^\circ\text{C} \) in a container, the resulting temperature is
1. \( 2.7^\circ\text{C} \)
2. \( 3.7^\circ\text{C} \)
3. \( 5.7^\circ\text{C} \)
4. \( 0^\circ\text{C} \)
View Answer
Heat lost by water to cool to \( 0^\circ\text{C} \) is \( Q_{\text{lost}} = m_w c_w \Delta T = 60 \times 1 \times 50 = 3000 \text{ cal} \). Heat needed to melt all ice is \( Q_{\text{melt}} = m_i L = 150 \times 80 = 12000 \text{ cal} \). Since \( Q_{\text{lost}} < Q_{\text{melt}} \), only a part of the ice melts, and the final temperature remains \( 0^\circ\text{C} \).
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 \(20^\circ\text{C}\))
1. 6 minutes
2. 10 minutes
3. 9 minutes
4. 4 minutes
View Answer
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, \(\frac{20}{t} = \frac{1}{9}[50 - 20] = \frac{30}{9} = \frac{10}{3} ⇒ t = 6 \text{ minutes}\).
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
1. \(\frac{7}{5}\)
2. \(\frac{7}{6}\)
3. \(\frac{4}{3}\)
4. \(\frac{3}{2}\)
View Answer
Total degrees of freedom \(f = 3 + 3 = 6\). Thus, \(C_v = \frac{f}{2} R = 3R\), and \(C_p = C_v + R = 4R\). Therefore, \(\gamma = \frac{C_p}{C_v} = \frac{4}{3}\).