Correct answer Carries: 4.
Wrong Answer Carries: -1.
What is the molar specific heat capacity at constant pressure for a diatomic gas if \( C_v = 20.75 \, \text{J mol}^{-1} \text{K}^{-1} \) and \( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \)?
\( C_p - C_v = R \).
\( C_p = C_v + R = 20.75 + 8.3 = 29.05 \, \text{J mol}^{-1} \text{K}^{-1} \approx 29.1 \, \text{J mol}^{-1} \text{K}^{-1} \).
A gas undergoes an adiabatic expansion from \( 25 \, \text{L} \) to \( 100 \, \text{L} \), reducing its pressure from \( 16 \, \text{atm} \) to \( 1 \, \text{atm} \). What is the value of \( \gamma \)?
For adiabatic: \( P_1 V_1^\gamma = P_2 V_2^\gamma \).
\( 16 \times 25^\gamma = 1 \times 100^\gamma \).
\( 16 = \left(\frac{100}{25}\right)^\gamma \Rightarrow 16 = 4^\gamma \).
\( 4^\gamma = 2^4 \Rightarrow 2^{2\gamma} = 2^4 \Rightarrow 2\gamma = 4 \Rightarrow \gamma = 2 \).
An ideal gas expands isothermally at \( 460 \, \text{K} \) from \( 3 \, \text{L} \) to \( 9 \, \text{L} \) with \( 0.5 \, \text{moles} \). What is the work done by the gas? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \))
For isothermal: \( W = \mu R T \ln\left(\frac{V_2}{V_1}\right) \).
\( \mu = 0.5 \), \( R = 8.3 \), \( T = 460 \), \( V_2 = 9 \), \( V_1 = 3 \).
\( W = 0.5 \times 8.3 \times 460 \times \ln\left(\frac{9}{3}\right) = 1909 \times \ln(3) \).
\( \ln(3) \approx 1.0986 \), \( W \approx 1909 \times 1.0986 \approx 2097 \, \text{J} \).
A system in a cyclic process absorbs \( 900 \, \text{J} \) of heat and rejects \( 400 \, \text{J} \). What is the net work done?
For cyclic process: \( \Delta U = 0 \), \( W = Q_{\text{net}} \).
\( Q_{\text{net}} = Q_{\text{absorb}} - Q_{\text{reject}} = 900 - 400 = 500 \, \text{J} \).
\( W = 500 \, \text{J} \).
How much heat is required to raise the temperature of \( 1 \, \text{kg} \) of copper from \( 20^\circ \text{C} \) to \( 50^\circ \text{C} \)? (Specific heat of copper = \( 386.4 \, \text{J kg}^{-1} \text{K}^{-1} \))
Heat capacity: \( \Delta Q = m s \Delta T \).
\( m = 1 \, \text{kg} \), \( s = 386.4 \, \text{J kg}^{-1} \text{K}^{-1} \), \( \Delta T = 50 - 20 = 30 \, \text{K} \).
\( \Delta Q = 1 \times 386.4 \times 30 = 11592 \, \text{J} \).
A gas undergoes a cyclic process where \( 600 \, \text{J} \) of heat is absorbed. What is the net work done by the gas?
For a cyclic process, \( \Delta U = 0 \).
\( \Delta Q = \Delta U + \Delta W \Rightarrow 600 = 0 + \Delta W \).
\( \Delta W = 600 \, \text{J} \).
A system in a cyclic process absorbs \( 1020 \, \text{J} \) of heat and rejects \( 380 \, \text{J} \). What is the net work done?
For cyclic: \( \Delta U = 0 \), \( Q_{\text{net}} = W \).
\( Q_{\text{net}} = Q_{\text{absorb}} - Q_{\text{reject}} = 1020 - 380 = 640 \, \text{J} \).
\( W = 640 \, \text{J} \).
What is the molar specific heat capacity at constant pressure for a solid if its molar specific heat is \( 24.4 \, \text{J mol}^{-1} \text{K}^{-1} \) at constant volume?
For solids, \( C \approx 3R \), but here \( C_v = 24.4 \), and \( \Delta V \approx 0 \), so \( C_p \approx C_v \).
However, typically \( C_p - C_v = R \), but for solids in PDF context, \( C \) is given directly.
Since \( C = 24.4 \) is molar specific heat, \( C_p \approx C_v = 24.4 \, \text{J mol}^{-1} \text{K}^{-1} \) (negligible \( \Delta V \)).
Correction: \( C_p \) not explicitly defined differently, use given value as per context: \( 24.4 \).
How much heat is required to raise the temperature of \( 0.6 \, \text{kg} \) of silver from \( 20^\circ \text{C} \) to \( 50^\circ \text{C} \)? (Specific heat of silver = \( 236.1 \, \text{J kg}^{-1} \text{K}^{-1} \))
\( \Delta Q = m s \Delta T \).
\( m = 0.6 \), \( s = 236.1 \), \( \Delta T = 50 - 20 = 30 \).
\( \Delta Q = 0.6 \times 236.1 \times 30 = 4249.8 \, \text{J} \approx 4250 \, \text{J} \).
What is the change in internal energy for \( 0.5 \, \text{moles} \) of an ideal gas heated from \( 250 \, \text{K} \) to \( 300 \, \text{K} \) at constant volume? (\( C_v = 20.8 \, \text{J mol}^{-1} \text{K}^{-1} \))
\( \Delta U = \mu C_v \Delta T \).
\( \mu = 0.5 \), \( C_v = 20.8 \), \( \Delta T = 300 - 250 = 50 \).
\( \Delta U = 0.5 \times 20.8 \times 50 = 520 \, \text{J} \).
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