Question

A cylinder contains 0.250 mol of carbon dioxide (CO2) gas at a temperature of 27.0∘C. The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 atm on the gas. The gas is heated until its temperature increases to 127.0∘C. Assume that the CO2 may be treated as an ideal gas. How much work W is done by the gas in this process? What is the change in internal energy ΔU of the gas? How much heat Q was supplied to the gas?

Answer 1

Concepts and reason

The concepts required to solve this problem are work done in an isobaric process, ideal gas law, internal energy and the first law of thermodynamics.

The provided temperatures are to be converted into Kelvin and then they are to be used to calculate the work done. Then the part on which work is done is to be mentioned. Further change in internal energy of the system is to be calculated. Then the heat supplied to the system is to be calculated. After that, the work done for the new value of pressure is to be determined.

Fundamentals

Isobaric Process: The process in which pressure remains constant is called isobaric process.

Ideal Gas Law: It provides a relationship between various measurable characteristics of a gas.

Internal Energy: It is the energy stored within the system. It is the sum of kinetic energy and the potential energy of the system.

First Law of Thermodynamics: First law of the thermodynamics states that the energy

can neither be created nor can be destroyed. It states that the energy interaction of a

system during a cycle is always constant.

Write the expression for work done in an isobaric process.

$W = P\left( {{V_2} - {V_1}} \right)$

Here, $P$ is the pressure, $W$ is work done, ${V_2}$ is the final volume and ${V_1}$ is the initial volume.

Write the expression for ideal gas law.

$PV = nRT$

Here, $n$ is number of moles, $V$ is volume, $T$ is temperature and $R$ is the universal gas constant.

Write the expression for change in internal energy.

$\Delta U = n{C_v}\left( {{T_2} - {T_1}} \right)$

Here, $\Delta U$ is change in internal energy, ${C_v}$ is specific heat at constant volume, ${T_2}$ is the final temperature and ${T_1}$ is the initial temperature.

Write the expression for first law of thermodynamics.

$Q = \Delta U + W$

Here, $Q$ is heat supplied.

(A)

Convert the provider temperatures from ${\rm{^\circ C}}$ to ${\rm{K}}$ by the following relation:

${T_K} = {T_C} + 273\;{\rm{K}}$ …… (1)

Substitute ${\rm{27 K}}$ for ${T_C}$ and ${{\rm{T}}_1}$ for ${{\rm{T}}_K}$ in equation (1).

$\begin{array}{c}\\{T_1} = 27\;{\rm{K}} + 273\;{\rm{K}}\\\\ = 300\;{\rm{K}}\\\end{array}$

Substitute ${\rm{127 K}}$ for ${T_C}$ and ${{\rm{T}}_2}$ for ${{\rm{T}}_K}$ in equation (1).

$\begin{array}{c}\\{T_2} = 127\;{\rm{K}} + 273\;{\rm{K}}\\\\ = 400\;{\rm{K}}\\\end{array}$

Write the expression for work done in an isobaric process.

$W = P\left( {{V_2} - {V_1}} \right)$ …… (2)

Consider gas law equation.

$PV = nRT$

$P\left( {{V_2} - {V_1}} \right) = nR\left( {{T_2} - {T_1}} \right)$ …… (3)

Substitute this value in equation (2).

$W = nR\left( {{T_2} - {T_1}} \right)$ …… (4)

Calculate the work done.

Substitute ${\rm{0}}{\rm{.250 mol}}$ for $n$ , ${\rm{8}}{\rm{.314 J/mol}} \cdot {\rm{K}}$ for ${\rm{R}}$ , ${\rm{400 K}}$ for ${{\rm{T}}_2}$ and ${\rm{300 K}}$ for ${{\rm{T}}_1}$ in equation (4).

$\begin{array}{c}\\W = \left( {{\rm{0}}{\rm{.250 mol}}} \right)\left( {{\rm{8}}{\rm{.314 J/mol}} \cdot {\rm{K}}} \right)\left( {{\rm{400 K}} - {\rm{300 K}}} \right)\\\\ = 207.9{\rm{ J}}\\\end{array}$

(B)

As the heat is supplied to the system at constant pressure, the temperature will increase and the gas will expand. The piston will move up. Hence, the work is done on the piston. There is no work on the cylinder as it is stationary.

‎

(C)

Calculate the change in internal energy of the system.

$\Delta U = n{C_v}\left( {{T_2} - {T_1}} \right)$

Substitute $0.250{\rm{ mol}}$ for $n$ , ${\rm{28}}{\rm{.46 J/mol}} \cdot {\rm{K}}$ for ${C_v}$ , ${\rm{400 K}}$ for ${{\rm{T}}_2}$ and ${\rm{300 K}}$ for ${{\rm{T}}_1}$ .

$\begin{array}{c}\\\Delta U = \left( {0.250{\rm{ mol}}} \right)\left( {{\rm{28}}{\rm{.46 J/mol}} \cdot {\rm{K}}} \right)\left( {{\rm{400 K}} - {\rm{300 K}}} \right)\\\\ = 711.5{\rm{ J}}\\\end{array}$

(D)

Heat supplied to the gas can be calculated by the first law of thermodynamics.

$Q = \Delta U + W$

Substitute $711.5{\rm{ J}}$ for $\Delta U$ and ${\rm{207}}{\rm{.9 J}}$ for $W$ .

$\begin{array}{c}\\Q = 711.5{\rm{ J}} + {\rm{207}}{\rm{.9 J}}\\\\{\rm{ = 919}}{\rm{.4 J}}\\\end{array}$

(E)

Calculate the work done.

$W = nR\left( {{T_2} - {T_1}} \right)$

Substitute ${\rm{0}}{\rm{.250 mol}}$ for $n$ , ${\rm{8}}{\rm{.314 J/mol}} \cdot {\rm{K}}$ for ${\rm{R}}$ , ${\rm{400 K}}$ for ${{\rm{T}}_2}$ and ${\rm{300 K}}$ for ${{\rm{T}}_1}$ .

$\begin{array}{c}\\W = \left( {{\rm{0}}{\rm{.250 mol}}} \right)\left( {{\rm{8}}{\rm{.314 J/mol}} \cdot {\rm{K}}} \right)\left( {{\rm{400 K}} - {\rm{300 K}}} \right)\\\\ = 207.9{\rm{ J}}\\\end{array}$

The work done will remain the same.

Ans: Part A

The work done is $207.9{\rm{ J}}$ .

Part B

The work is done on the piston.

Part C

The change in internal energy of the system is $711.5{\rm{ J}}$ .

Part D

The heat supplied to the gas is ${\rm{919}}{\rm{.4 J}}$ .

Part E

The work done is $207.9{\rm{ J}}$ .