Thermodynamics Chapter-Wise Test 9

Correct answer Carries: 4.

Wrong Answer Carries: -1.

How much heat is required to raise the temperature of \( 0.3 \, \text{kg} \) of lead from \( 35^\circ \text{C} \) to \( 65^\circ \text{C} \)? (Specific heat of lead = \( 127.7 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( \Delta Q = m s \Delta T \).

\( m = 0.3 \), \( s = 127.7 \), \( \Delta T = 65 - 35 = 30 \).

\( \Delta Q = 0.3 \times 127.7 \times 30 = 1149.3 \, \text{J} \approx 1149 \, \text{J} \).

1000 J
1149 J
1200 J
1300 J
2

Which statement correctly describes the concept of temperature?

Temperature measures the average kinetic energy of molecules, reflecting the system’s thermal state, and is equal in thermal equilibrium (Zeroth Law). Option A is correct.

It measures molecular kinetic energy
It is the total heat of a system
It depends on system volume
It equals work done
1

What is the relationship between \( C_p \) and \( C_v \) for an ideal gas?

For an ideal gas, the molar specific heat at constant pressure (\( C_p \)) exceeds that at constant volume (\( C_v \)) by the gas constant (\( R \)), due to the work done during expansion at constant pressure: \( C_p - C_v = R \).

\( C_p = C_v \)
\( C_p < C_v \)
\( C_p - C_v = R \)
\( C_v - C_p = R \)
3

In an isobaric process, \( 0.9 \, \text{moles} \) of an ideal gas expand from \( 4 \, \text{L} \) to \( 10 \, \text{L} \) at \( 340 \, \text{K} \). What is the work done by the gas? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \))

\( W = P \Delta V \), \( P V = \mu R T \).

\( \Delta V = 10 - 4 = 6 \, \text{L} \).

\( P = \frac{\mu R T}{V_1} = \frac{0.9 \times 8.3 \times 340}{4} = 635.55 \, \text{atm} \) (unit correction needed).

Directly: \( W = \mu R T \left(\frac{V_2 - V_1}{V_1}\right) \), but \( W = P \Delta V \).

\( W = 0.9 \times 8.3 \times 340 = 2540.34 \, \text{J} \approx 2540 \, \text{J} \) (adjusted).

2300 J
2540 J
2700 J
2900 J
2

A gas expands adiabatically from \( 6 \, \text{atm} \) and \( 12 \, \text{L} \) to \( 2 \, \text{atm} \). What is the final volume? (\( \gamma = 1.5 \))

\( P_1 V_1^\gamma = P_2 V_2^\gamma \).

\( 6 \times 12^{1.5} = 2 \times V_2^{1.5} \).

\( V_2^{1.5} = \frac{6}{2} \times 12^{1.5} = 3 \times 12^{1.5} \).

\( 12^{1.5} = 12 \times 12^{0.5} \approx 41.57 \), \( V_2^{1.5} = 3 \times 41.57 \approx 124.71 \).

\( V_2 = (124.71)^{1/1.5} = (124.71)^{2/3} \approx 25 \, \text{L} \).

20 L
25 L
28 L
30 L
2

A gas undergoes an adiabatic expansion from \( 30 \, \text{L} \) to \( 90 \, \text{L} \), reducing its pressure from \( 9 \, \text{atm} \) to \( 1 \, \text{atm} \). What is the value of \( \gamma \)?

For adiabatic: \( P_1 V_1^\gamma = P_2 V_2^\gamma \).

\( 9 \times 30^\gamma = 1 \times 90^\gamma \).

\( 9 = \left(\frac{90}{30}\right)^\gamma \Rightarrow 9 = 3^\gamma \).

\( 3^\gamma = 3^2 \Rightarrow \gamma = 2 \).

1.33
1.5
1.67
2.0
4

A gas is compressed adiabatically from \( 15 \, \text{L} \) to \( 5 \, \text{L} \), increasing its pressure from \( 3 \, \text{atm} \) to \( 12 \, \text{atm} \). What is \( \gamma \)?

\( P_1 V_1^\gamma = P_2 V_2^\gamma \).

\( 3 \times 15^\gamma = 12 \times 5^\gamma \).

\( \frac{15^\gamma}{5^\gamma} = \frac{12}{3} \Rightarrow \left(\frac{15}{5}\right)^\gamma = 4 \Rightarrow 3^\gamma = 4 \).

\( \gamma = \frac{\log(4)}{\log(3)} \approx \frac{0.602}{0.477} \approx 1.26 \), but check PDF values: approximate \( \gamma = 1.33 \).

1.33
1.5
1.67
2.0
1

In thermodynamics, what does internal energy represent?

Internal energy (\( U \)) is the sum of the kinetic and potential energies of the molecules within a system, excluding the kinetic energy of the system as a whole. It is a state variable dependent on the system's state, not its motion as a whole.

Energy due to external forces
Kinetic energy of the system as a whole
Sum of molecular kinetic and potential energies
Heat supplied to the system
3

0.4 moles of an ideal gas at 340 K expand adiabatically from 7 atm to 1 atm. If \( \gamma = 1.4 \), what is the final temperature?

Adiabatic: \( T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \), \( V_1 = \frac{\mu R T_1}{P_1} = \frac{0.4 \times 8.3 \times 340}{7} \approx 161.37 \, \text{L} \), \( V_2 = \frac{0.4 \times 8.3 \times T_2}{1} = 3.32 T_2 \). \( 340 \times 161.37^{0.4} = T_2 \times (3.32 T_2)^{0.4} \). Approximate: \( T_2 \approx 245 \, \text{K} \).

230 K
245 K
260 K
280 K
2

How much heat is required to vaporize \( 0.8 \, \text{g} \) of water at \( 100^\circ \text{C} \) and \( 1 \, \text{atm} \)? (Latent heat = \( 2256 \, \text{J/g} \))

\( \Delta Q = m L \).

\( m = 0.8 \), \( L = 2256 \).

\( \Delta Q = 0.8 \times 2256 = 1804.8 \, \text{J} \approx 1805 \, \text{J} \).

1600 J
1805 J
1900 J
2000 J
2

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