The internal energy of an ideal monoatomic gas increases by the same amount as work done on the gas, then:
By the first law, \(dQ = dU + dW\). If work is done on the gas, \(dW_{\text{by}} = -dW_{\text{on}}\). Given \(dU = dW_{\text{on}}\), so \(dQ = 0\). This represents an adiabatic process.
The work done by 3 moles of gas at 47°C to triple its volume at constant pressure is (\(R = 2\) cal mol\(^{-1}\) °C\(^{-1}\)):
At constant pressure, \(W = nR\Delta T\). Since volume triples, temperature also triples (from \(320\) K to \(960\) K), so \(\Delta T = 640\) K. Work \(W = 3 \times 2 \times 640 = 3840\) cal.
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
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.
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.
Two moles of an ideal monoatomic gas undergoes an adiabatic process from temperature \(300\text{ K}\) to \(600\text{ K}\). Work done by this ideal gas in the process is
Work done in an adiabatic process is \(W = \frac{nR(T_1 - T_2)}{\gamma - 1}\). For monoatomic gas, \(\gamma = 5/3\). Substituting the parameters: \(W = \frac{2R(300 - 600)}{5/3 - 1} = \frac{-600R}{2/3} = -900R\).
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
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
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 \( 0^\circ\text{C} \).
1 g of steam at 100°C is mixed with 1 g of ice at 0°C, then resultant temperature of the mixture is
Heat required to raise 1 g of ice at 0°C to water at 100°C is \(1 \times 80 + 1 \times 1 \times 100 = 180\text{ cal}\). Since 1 g steam releases \(540\text{ cal}\) upon complete condensation, only a fraction of the steam condenses, maintaining the final temperature at 100°C.
110 J of heat is added to a gaseous system whose internal energy is increased by 40 J then amount of external work done is
According to the first law of thermodynamics, \(\Delta Q = \Delta U + W\). Substituting the values, \(110\text{ J} = 40\text{ J} + W \Rightarrow W = 70\text{ J}\).
At 10°C the value of the density of a fixed mass of an ideal gas divided by its pressure is X. At 110°C this ratio is
From ideal gas law, \(P = \frac{\rho RT}{M} \Rightarrow \frac{\rho}{P} = \frac{M}{RT}\). Hence, \(\frac{\rho}{P} \propto \frac{1}{T}\). Thus, the ratio becomes \[X \times \frac{273 + 10}{273 + 110} = X \frac{283}{383}\].