**30 Application of the first law of thermodynamics in some thermodynamic processes (Isobaric Isothermal Isochoric)**

1. The graph below shows the thermodynamic cycle experienced by a gas. The work done by gas on the process ABCD is …

__Known :__

Pressure 1 (P_{1}) = 2 x 10^{5} Pa

Pressure 2 (P_{2}) = 4 x 10^{5} Pa

Volume 1 (V_{1}) = 1 m^{3}

Volume 2 (V_{2}) = 3 m^{3}

Wanted: Work done by gas on process ABCD (W).

__Solution :__

Work done by a gas is equal to the area of ABCD.

W = (P_{2} – P_{1})(V_{2} – V_{1})

W = (4 x 10^{5} – 2 x 10^{5})(3 – 1)

W = (2 x 10^{5})(2)

W = 4 x 10^{5} Joule

2. The work done by gas on process ABC is…

__Known :__

Pressure 1 (P_{1}) = 3 x 10^{5} Pa

Pressure 2 (P_{2}) = 6 x 10^{5} Pa

Volume 1 (V_{1}) = 20 cm^{3} = 20 x 10^{-6} m^{3}

Volume 2 (V_{2}) = 60 cm^{3} = 60 x 10^{-6} m^{3}

__Wanted __: The work done by gas on process ABC

__Solution :__

Work done by gas = area of ABC.

W = ½ (P_{2} – P_{1})(V_{2} – V_{1})

W = 1/2 (6 x 10^{5} – 3 x 10^{5})(60 x 10^{-6} – 20 x 10^{-6})

W = 1/2 (3 x 10^{5})(40 x 10^{-6})

W = 1/2 (120 x 10^{-1} Joule)

W = 1/2 (12 Joule)

W = 6 Joule

3. PV diagram for a gas in a closed container shown in figure below. The work done by the gas is shown in which process.

Solution

Process AB and process DC are isobaric processes (the pressure is kept constant). Process AD and process BC are isochoric processes (volume does not change).

The work was done by gas when the gas expands (process DC).

Here are 9 problems and solutions related to the Isobaric Process (Constant Pressure):

Problem 1:

A 3.0 m³ gas at constant pressure is compressed to 2.0 m³. If the initial temperature is 300 K, what is the final temperature?

Solution:

Using the equation for an isobaric process: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), we can solve for the final temperature:

\[

\frac{3.0}{300} = \frac{2.0}{T_2} \implies T_2 = \frac{2.0 \times 300}{3.0} = 200\, \text{K}

\]

Problem 2:

Calculate the work done by 2 moles of gas expanding isobarically from 1 liter to 3 liters at 2 atm.

Solution:

Using the equation for work done in an isobaric process: \( W = P \Delta V \):

\[

W = 2\, \text{atm} \times (3 – 1)\, \text{liters} = 4\, \text{atm}\cdot\text{liters}

\]

Problem 3:

What is the change in internal energy when 4 moles of an ideal diatomic gas are heated at constant pressure from 200 K to 300 K?

Solution:

Using the equation for the change in internal energy at constant pressure for a diatomic gas: \( \Delta U = nC_pdT \), where \( C_p = \frac{7}{2}R \):

\[

\Delta U = 4 \times \frac{7}{2} \times 8.314 \times (300 – 200) \approx 6990\, \text{J}

\]

Problem 4:

Calculate the heat added during an isobaric process when 2 moles of a monatomic gas expand from 5 liters to 10 liters at 400 K.

Solution:

Using the equation for heat transfer in an isobaric process: \( Q = nC_pdT \), where \( C_p = \frac{5}{2}R \), and \( \frac{V_2}{T_2} = \frac{V_1}{T_1} \):

\[

Q = 2 \times \frac{5}{2} \times 8.314 \times \left(400 \times \frac{10}{5} – 400\right) = 3316\, \text{J}

\]

Problem 5:

Determine the change in enthalpy of 3 moles of a triatomic gas that is expanded from 1 liter to 2 liters at a temperature of 200 K.

Solution:

Using the formula for the change in enthalpy for an isobaric process: \( \Delta H = nC_pdT \), where \( C_p = \frac{f}{2}R \) and \( f = 6 \):

\[

\Delta H = 3 \times \frac{6}{2} \times 8.314 \times \left(200 \times \frac{2}{1} – 200\right) = 4989\, \text{J}

\]

Problem 6:

Calculate the final volume of a 2 moles of a gas at 300 K and 5 liters initially when it is heated to 500 K at constant pressure.

Solution:

Using the equation for an isobaric process:

\[

\frac{V_1}{T_1} = \frac{V_2}{T_2} \implies V_2 = \frac{V_1 \times T_2}{T_1} = \frac{5 \times 500}{300} \approx 8.33\, \text{liters}

\]

Problem 7:

Find the work done by a monatomic gas when it is compressed from 6 liters to 3 liters at 2 atm.

Solution:

Using the equation for work done in an isobaric process:

\[

W = P \Delta V = 2 \times (3 – 6) = -6\, \text{atm}\cdot\text{liters}

\]

Problem 8:

A diatomic gas undergoes an isobaric expansion from 3 liters to 6 liters at 300 K. Find the change in entropy.

Solution:

Using the equation for the change in entropy in an isobaric process for a diatomic gas, where \( C_p = \frac{7}{2}R \):

\[

\Delta S = nC_p\ln\frac{T_2}{T_1} + nR\ln\frac{V_2}{V_1} = 2 \times \frac{7}{2} \times 8.314 \times \ln\frac{300 \times \frac{6}{3}}{300} + 2 \times 8.314 \times \ln\frac{6}{3} \approx 34.76\, \text{J/K}

\]

Problem 9:

Calculate the heat transfer during an isobaric compression of 5 moles of a triatomic gas from 10 liters to 5 liters at 500 K.

Solution:

Using the equation for heat transfer in an isobaric process, where \( C_p = \frac{f}{2}R \) and \( f = 6 \):

\[

Q = 5 \times \frac{6}{2} \times 8.314 \times \left(500 \times \frac{5}{10} – 500\right) = -12473\, \text{J}

\]

These problems should give a comprehensive overview of various concepts related to the isobaric process.

Here are 9 problems and solutions related to the Isothermal Process (Constant Temperature):

Problem 1:

Problem: Calculate the work done by 2 moles of an ideal gas expanding isothermally from 1 liter to 2 liters at 300 K.

Solution:

Using the equation for work done in an isothermal process for an ideal gas:

\[ W = nRT\ln\frac{V_2}{V_1} \]

\[

W = 2 \times 8.314 \times 300 \times \ln\frac{2}{1} \approx 3454\, \text{J}

\]

Problem 2:

Problem: Find the heat transferred when 3 moles of a gas are compressed isothermally from 4 liters to 2 liters at 400 K.

Solution:

Using the equation for heat transfer in an isothermal process: \( Q = W \):

\[

Q = 3 \times 8.314 \times 400 \times \ln\frac{2}{4} \approx -3462\, \text{J}

\]

Problem 3:

Determine the change in entropy when 4 moles of a gas undergo an isothermal expansion from 1 liter to 5 liters at 300 K.

Solution:

Using the equation for the change in entropy in an isothermal process:

\[

\Delta S = nR\ln\frac{V_2}{V_1} = 4 \times 8.314 \times \ln\frac{5}{1} \approx 46.15\, \text{J/K}

\]

Problem 4:

Calculate the work done when 1 mole of an ideal gas is compressed isothermally from 6 liters to 3 liters at 200 K.

Solution:

Using the equation for work done in an isothermal process:

\[

W = 1 \times 8.314 \times 200 \times \ln\frac{3}{6} \approx -575\, \text{J}

\]

Problem 5:

Determine the change in internal energy for 3 moles of gas during an isothermal expansion from 2 liters to 6 liters at constant temperature.

Solution:

For an isothermal process of an ideal gas, the change in internal energy is zero:

\[

\Delta U = 0\, \text{J}

\]

Problem 6:

Calculate the heat transfer when 5 moles of a gas are expanded isothermally from 3 liters to 6 liters at 250 K.

Solution:

Using the equation for heat transfer in an isothermal process:

\[

Q = 5 \times 8.314 \times 250 \times \ln\frac{6}{3} \approx 2879\, \text{J}

\]

Problem 7:

What is the work done by 2 moles of a gas during an isothermal compression from 4 liters to 2 liters at 150 K?

Solution:

Using the equation for work done in an isothermal process:

\[

W = 2 \times 8.314 \times 150 \times \ln\frac{2}{4} \approx -1151\, \text{J}

\]

Problem 8:

Find the change in entropy when 3 moles of a gas are compressed isothermally from 5 liters to 1 liter at 500 K.

Solution:

Using the equation for the change in entropy in an isothermal process:

\[

\Delta S = 3 \times 8.314 \times \ln\frac{1}{5} \approx -34.77\, \text{J/K}

\]

Problem 9:

Determine the work done by 4 moles of an ideal gas expanding isothermally from 2 liters to 8 liters at 100 K.

Solution:

Using the equation for work done in an isothermal process:

\[

W = 4 \times 8.314 \times 100 \times \ln\frac{8}{2} \approx 2304\, \text{J}

\]

These problems cover various concepts related to the isothermal process, such as work done, heat transfer, and change in entropy.

Here are 9 problems and solutions related to the Isochoric Process (Constant Volume):

Problem 1:

Calculate the change in internal energy for 3 moles of a monatomic ideal gas when the temperature is increased from 200 K to 400 K at constant volume.

Solution:

Using the equation for change in internal energy at constant volume: \( \Delta U = nC_v\Delta T \), where \( C_v = \frac{3}{2}R \):

\[

\Delta U = 3 \times \frac{3}{2} \times 8.314 \times (400 – 200) \approx 3741\, \text{J}

\]

Problem 2:

Determine the heat transfer for 2 moles of a diatomic gas when the temperature is decreased from 300 K to 200 K at constant volume.

Solution:

Using the equation for heat transfer at constant volume: \( Q = \Delta U = nC_v\Delta T \), where \( C_v = \frac{5}{2}R \):

\[

Q = 2 \times \frac{5}{2} \times 8.314 \times (200 – 300) \approx -4157\, \text{J}

\]

Problem 3:

Calculate the change in entropy for 4 moles of a triatomic gas when the temperature increases from 100 K to 300 K at constant volume.

Solution:

Using the equation for the change in entropy at constant volume, where \( C_v = \frac{f}{2}R \) and \( f = 6 \):

\[

\Delta S = nC_v\ln\frac{T_2}{T_1} = 4 \times \frac{6}{2} \times 8.314 \times \ln\frac{300}{100} \approx 115.36\, \text{J/K}

\]

Problem 4:

What is the work done by a gas during an isochoric process?

Solution:

Since the volume is constant during an isochoric process, there is no work done:

\[

W = 0\, \text{J}

\]

Problem 5:

Determine the heat transfer for 5 moles of a monatomic ideal gas when it is cooled from 500 K to 300 K at constant volume.

Solution:

Using the equation for heat transfer at constant volume, where \( C_v = \frac{3}{2}R \):

\[

Q = 5 \times \frac{3}{2} \times 8.314 \times (300 – 500) \approx -6232\, \text{J}

\]

Problem 6:

Calculate the change in internal energy for 1 mole of a diatomic gas when the temperature is increased from 150 K to 250 K at constant volume.

Solution:

Using the equation for change in internal energy at constant volume, where \( C_v = \frac{5}{2}R \):

\[

\Delta U = 1 \times \frac{5}{2} \times 8.314 \times (250 – 150) \approx 2079\, \text{J}

\]

Problem 7:

Determine the change in entropy for 3 moles of a monatomic gas when the temperature decreases from 600 K to 300 K at constant volume.

Solution:

Using the equation for the change in entropy at constant volume, where \( C_v = \frac{3}{2}R \):

\[

\Delta S = 3 \times \frac{3}{2} \times 8.314 \times \ln\frac{300}{600} \approx -34.59\, \text{J/K}

\]

Problem 8:

What is the heat transfer for 2 moles of a triatomic gas when the temperature is increased from 200 K to 400 K at constant volume?

Solution:

Using the equation for heat transfer at constant volume, where \( C_v = \frac{6}{2}R \):

\[

Q = 2 \times \frac{6}{2} \times 8.314 \times (400 – 200) \approx 4986\, \text{J}

\]

Problem 9:

Calculate the change in internal energy for 4 moles of a monatomic ideal gas when it is heated from 250 K to 350 K at constant volume.

Solution:

Using the equation for change in internal energy at constant volume, where \( C_v = \frac{3}{2}R \):

\[

\Delta U = 4 \times \frac{3}{2} \times 8.314 \times (350 – 250) \approx 4986\, \text{J}

\]

These problems cover various aspects of the isochoric process, such as heat transfer, changes in internal energy, and entropy, for different types of gases.