Question 1

A gas can absorb heat without changing temperature if at the same time

Accepted Answer

The answer of A gas can absorb heat without changing...

Question 2

The internal energy for a diatomic gas is given by U = 5nRT/2. Calculate the internal energy of a 100 g mixture of oxygen 20%) and nitrogen 80%) gas at 25°C. The molar weight of O₂ = 32 g, and the molar weight of N₂ = 28 g.)

Accepted Answer

The total moles of the gas mixture are calculated using the given mass percentages and molar weights. Then, using U = 5nRT/2 with R = 8.314 J/mol·K and T = 298 K, the internal energy is calculated.

Question 3

The specific heat of a gas is

Accepted Answer

The specific heat of a gas is greater at constant pressure than at constant volume because when a gas is heated at constant pressure, work is done by the gas as it expands, requiring more heat energy for a given temperature rise compared to heating at constant volume where no work is done.

Question 4

The diagram above show the state of an ideal gas going from V₁, P₁) to a final state. Which path best represents an isothermal expansion?

Accepted Answer

The process of isothermal expansion occurs at constant temperature. Path 2 represents an isothermal expansion because it stays at a constant temperature while the volume increases, indicating that the work done is being used to increase internal energy rather than increase temperature.

Question 5

One mole of an ideal gas γ = 5/3) expands adiabatically and quasi-statically from a pressure P₁ = 3 atm and a temperature of 30^oC to a pressure P₂ = 1 atm. How much work is done by the gas during this process? R = 8.314 J/mol · K = 8.206 L · atm/mol · K.

Accepted Answer

The answer of One mole of an ideal gas γ...

Question 6

For an ideal gas, the difference in the molar heat capacity at constant P and constant V is

Accepted Answer

For an ideal gas, the difference in the molar heat capacity at constant P and constant V is given by the Mayer's formula: Cp - Cv = R, where Cp is the molar heat capacity at constant pressure, Cv is the molar heat capacity at constant volume, and R is the gas constant. This equation holds true for all gases, regardless of whether they are monoatomic, diatomic, or polyatomic, as long as they can be approximated as an ideal gas. Therefore, the best choice is E) A) and D), which states that the difference is equal to R for all gases.

Question 7

The pressure of a mass of air at 20°C is halved adiabatically. If the ratio of C_p to C_v for air is 1.41, calculate the resulting volume.

Accepted Answer

According to adiabatic process formula, $P_1V_1^{\gamma}=P_2V_2^{\gamma}$ where $\gamma=\frac{C_p}{C_v}$ for the air is 1.41.
If the pressure is halved, then the new pressure becomes $\frac{1}{2}P_1$. The formula becomes: 
$$\frac{1}{2}P_1V_1^{\gamma} = P_1 (\frac{V_2}{V_1})^{\gamma} V_2^{\gamma}$$

simplifies to: $$\frac{1}{2} = (\frac{V_2}{V_1})^{\gamma}$$

$$V_2 = V_1(\frac{1}{2})^{\frac{1}{\gamma}}$$

$$V_2 = V_1(\frac{1}{2})^{\frac{1}{1.41}}$$

$$V_2 = 1.63 V_1$$

Therefore, the resulting volume is 1.63 times the original volume. The correct answer is B.

Question 8

A gas has a molar heat capacity at constant volume of 28.39 J/mol · K. Assume the equipartition theorem to be valid. How many degrees of freedom including translational) are there for the molecules of this gas? The ideal-gas law constant is R = 8.31 J/mol · K)

Accepted Answer

The molar heat capacity at constant volume, Cv, is given by (f/2)R, where f is the degrees of freedom. Solving 28.39 J/mol·K = (f/2)(8.31 J/mol·K) gives f = 7.

Question 9

One mole of an ideal gas γ = 5/3) expands adiabatically and quasi-statically from a pressure P₁ = 6 atm and a temperature of 50^oC to a pressure P₂ = 4 atm. How much work is done by the gas during this process? R = 8.314 J/mol · K = 8.206 L · atm/mol · K.

Accepted Answer

We can use the adiabatic work formula: $W=\frac{\gamma}{\gamma - 1}P_1V_1\left(1-\left(\frac{P_2}{P_1}ight)^{\frac{\gamma-1}{\gamma}}ight)$
Using the ideal gas law, we can find that $V_1=\frac{nRT_1}{P_1}=9.12$ L and $V_2=\frac{nRT_2}{P_2}=12.18$ L. 
Substituting the given values into the adiabatic work formula, we get 
$W=\frac{5}{3-1}(6	ext{ atm}\cdot9.12	ext{ L})\left(1-\left(\frac{4}{6}ight)^\frac{2}{5}ight)=56.2$ kJ.

Question 10

An ideal gas initially at 50^oC and pressure P₁ = 100 kPa occupies a volume V₁ = 3 L. It undergoes a quasi-static, isothermal expansion until its pressure is reduced to 50 kPa. How much heat enters the gas during this process? R = 8.314 J/mol · K = 8.206 L · atm/mol · K.

Accepted Answer

The answer of An ideal gas initially at 50^oC and...

Question 11

The specific heat of a gas at constant pressure is

Accepted Answer

The specific heat of a gas at constant pressure (Cp) is always greater than the specific heat at constant volume (Cv) due to the work done by the gas during expansion at constant pressure.

Question 12

The diagram above shows the state of an ideal gas going from V₁,P₁) to a final state. Which path best represents adiabatic expansion?

Accepted Answer

The answer of  The diagram above shows the state...

Question 13

The molar heat capacity at constant volume of a gas is found to be 20.74 J/mol · K. What is the molar heat capacity at constant pressure of this gas? The ideal-gas law constant is R = 8.31 J/mol · K.)

Accepted Answer

The relationship between molar heat capacity at constant pressure (Cp) and molar heat capacity at constant volume (Cv) for an ideal gas is:

Cp - Cv = R

where R is the ideal-gas law constant.

Rearranging this equation, we can solve for Cp:

Cp = Cv + R

Plugging in the given value of Cv (20.74 J/mol · K) and R (8.31 J/mol · K), we get:

Cp = 20.74 J/mol · K + 8.31 J/mol · K = 29.05 J/mol · K

Therefore, the molar heat capacity at constant pressure of this gas is approximately 29.0 J/mol · K, which is closest to choice B.

Question 14

From the measured molar heat capacities and the equipartition theorem, for a polyatomic gas molecule the number of degrees of freedom due to translational motion are

Accepted Answer

According to the equipartition theorem, each molecule in a polyatomic gas has 3 degrees of freedom due to translational motion, corresponding to movement along the x, y, and z axes.

Question 15

In a system composed of an ideal gas contained in a cylinder fitted with a piston, a reversible adiabatic expansion causes the temperature of the gas to drop because

Accepted Answer

In a reversible adiabatic process, no heat is exchanged with the surroundings. The work done by the gas during expansion comes from its internal energy, leading to a decrease in temperature.

Question 16

From the measured molar heat capacities and the equipartition theorem, for a diatomic gas molecule the number of degrees of freedom from rotational motion are

Accepted Answer

The equipartition theorem tells us that each degree of freedom contributes 1/2R to the molar heat capacity. For a diatomic gas molecule, there are 3 degrees of freedom from translational motion and 2 degrees of freedom from rotational motion (since a diatomic molecule can rotate about two axes perpendicular to the bond axis), for a total of 5 degrees of freedom. Therefore, the molar heat capacity should be (5/2)R. However, experimental data shows that the molar heat capacity is actually slightly less than this, indicating that at low temperatures, some of the rotational degrees of freedom are not accessible due to quantum effects. This is known as the quantum mechanical "rotational heat capacity" and explains why the measured molar heat capacity is slightly less than expected.

Question 17

An ideal monatomic gas has a molar heat capacity C_mp at constant pressure. What is the molar heat capacity at constant volume of an ideal diatomic gas?

Accepted Answer

The answer of An ideal monatomic gas has a molar...

Question 18

The pressure of a mass of air at 20°C is halved adiabatically. If the ratio of C_p to C_v for air is 1.41, calculate the resulting temperature.

Accepted Answer

Since the process is adiabatic, we can use the adiabatic relation:
pV^γ = constant
where p is the pressure, V is the volume, and γ is the ratio of specific heats.
If the pressure is halved, then the final pressure is p/2. Since the volume V is not given, we can use the fact that the initial and final conditions are both on the same adiabat to relate the initial and final volumes:
(p/2)V^γ = pV^γ
Simplifying gives Vf = √2 Vi, where Vi is the initial volume and Vf is the final volume.
Now we can use the first law of thermodynamics, which for an adiabatic process reduces to:
PV^γ = constant
or
T = (P/V)^(1-γ) = constant * P^(γ-1)
where T is the temperature and P is the pressure.
Since the initial and final conditions are both on the same adiabat:
P1/2 * √2 Vi = P Vi
or
Pf = P/√2
Also, since the ratio of specific heats for air is 1.41, we have:
γ = 1.41
Plugging in all the values, we get:
Tf = Ti * (Pf/Pi)^(γ-1)
= 20°C * ((P/√2)/P)^(1.41-1)
= 20°C * (√2)^(-0.41)
= 20°C * 0.802
= 16.04°C
Therefore, the resulting temperature is approximately -33.0°C (Choice C).

Question 19

An ideal gas with an initial volume of 3 L at a pressure of 2 atm is compressed adiabatically until it has a volume of 2 L; then it is cooled at constant volume until its temperature drops to its initial value. The final pressure is

Accepted Answer

The answer of An ideal gas with an initial volume...

Question 20

Consider the following statement: "The specific heat of an ideal gas at constant pressure C_p is greater than the specific heat of a gas at constant volume C_v." Which of the following describes this statement?

Accepted Answer

The statement is true because, at constant pressure, the gas does work as it expands, requiring more energy input to achieve the same temperature change compared to constant volume, where no work is done.

Quiz 15: Thermodynamics II

A gas can absorb heat without changing temperature if at the same time

The internal energy for a diatomic gas is given by U = 5nRT/2. Calculate the internal energy of a 100 g mixture of oxygen 20%) and nitrogen 80%) gas at 25°C. The molar weight of O2 = 32 g, and the molar weight of N2 = 28 g.)

The specific heat of a gas is

The diagram above show the state of an ideal gas going from V1, P1) to a final state. Which path best represents an isothermal expansion?

One mole of an ideal gas γ = 5/3) expands adiabatically and quasi-statically from a pressure P1 = 3 atm and a temperature of 30oC to a pressure P2 = 1 atm. How much work is done by the gas during this process? R = 8.314 J/mol · K = 8.206 L · atm/mol · K.