The work done by the gas against the external pressure of [tex]10.00\text{ atm}[/tex] is [tex]\boxed{395.07\text{ J}}[/tex].
Further explanation:
This is a case of constant pressure expansion process. It is also called isobaric process. In an isobaric process gas does work at constant pressure.
According to first law of thermodynamics, for a process the net heat transfer to the system can be either increase the internal energy of the gas or do some work against an external force. If heat is added at constant pressure process, some heat is used in gas work against external pressure force and some of the work is used in increasing the internal energy of the gas.
Given:
Balloon is inflated from [tex]0.0100\text{ }l-0.400\text{ }l[/tex].
[tex]{V_1} = 0.0100\,{\text{ }l}\\{V_2} = 0.400\,{\text{ }l}[/tex]
External pressure is [tex]10.00\text{ atm}[/tex]
[tex]1\text{ atm}=101.3\text{ KPa}\\1\,{\text{}l} \times 1{\text{ atm}} = 101.3\,{\text{J}}[/tex]
Concept:
For a reversible process work done by gas is
[tex]dW = p \cdot dV[/tex]
Here, [tex]dW[/tex] is the work done in a process, [tex]p[/tex] is the pressure and [tex]dv[/tex] is the change in the volume of gas.
Integrating the above expression for a process 1 to 2
[tex]\int_1^2 {dW} = \int_1^2 {p \cdot dV}[/tex]
If pressure is kept constant in the process 1 to 2
Net work done in the process
[tex]{W_{{\text{net}}}} = p\left( {{V_2} - {V_1}} \right)[/tex] …… (1)
Substitute [tex]0.0100\,{\text{ }l}[/tex] for [tex]{V_1}[/tex], [tex]0.400\,{\text{ }l}[/tex] for [tex]{V_2}[/tex] and [tex]10.00\text{ atm}[/tex] for [tex]p[/tex] in equation (1).
[tex]\begin{aligned}{W_{{\text{net}}}}&=\left( {10.00\,{\text{atm}}} \right)\left( {0.400{\kern 1pt} {\text{l}} - 0.0100\,{\text{l}}} \right)\\&=3.90\,\,{\text{l}}\cdot{\text{atm}}\\\end{aligned}[/tex]
Use conversion
[tex]\begin{aligned}{W_{{\text{net}}}}&=3.90{\kern 1pt} {\kern 1pt} {\text{l}} \cdot {\text{atm}}\left( {\frac{{101.3\,{\text{J}}}}{{1\,\,{\text{l}} \cdot {\text{atm}}}}} \right)\\&=395.07\,{\text{J}}\\\end{gathered}[/tex]
Thus, the work done by the gas against the external pressure of [tex]10.00\text{ atm}[/tex] is [tex]\boxed{395.07\text{ J}}[/tex].
Learn More:
1. To what volume will a 2.33 l sample expand https://brainly.com/question/9979757
2. Kinetic energy of the emitted electrons https://brainly.com/question/9059731
Answer Details:
Grade: High School
Subject: Physics
Chapter: Thermodynamics
Keywords:
Balloon, inflated, 0.0100 l, 0.400 l, against, pressure, 10.00 atm, work, joules, 101.3 J, 1 ltimes atm, 3.9 l atm, 395.07 J.
