Respuesta :
a)
At constant pressure, the work done for gas to expand is given by:
W = PΔV
⇒W = 1.65 × 10⁵Pa × (0.320 - 0.110)m³= 3.46 × 10⁴J
b)
The total heat (Q) added to the system is used in work done and the remaining energy goes into internal energy.
ΔU = ΔQ-ΔW
⇒ΔU = 1.15 ×10⁵J - 3.46 × 10⁴J = 0.8 × 10⁵ J
c)
It does not matter that the gas is ideal or real because the work done would remain same. It would vary for real gases only at very high pressure.
(a). The work done by the gas is [tex]\boxed{3.47 \times {{10}^4}\,{\text{J}}}[/tex].
(b). The change in the internal energy of the gas is [tex]\boxed{8.03 \times {{10}^4}\,{\text{J}}}[/tex].
(c). There is no effect of the gas being ideal gas or real gas on the work done by the gas.
Further Explanation:
Given:
The initial volume of the gas is [tex]0.110\,{{\text{m}}^{\text{3}}}[/tex] .
The final volume of the gas is [tex]0.320\,{{\text{m}}^{\text{3}}}[/tex] .
The constant pressure at which the expansion of the gas takes place is [tex]1.65 \times {10^5}\,{\text{Pa}}[/tex] .
The amount of heat added to the gas is [tex]1.15 \times {10^5}\,{\text{J}}[/tex].
Concept:
Part (a):
The expansion of the gas is taking place at constant pressure. Therefore, the work done by the gas is:
[tex]W = P\left( {{V_f} - {V_i}}\right)[/tex]
Here, [tex]P[/tex] is the pressure and [tex]{V_f}[/tex] is the final volume and [tex]{V_i}[/tex] is the initial volume of the gas.
Substitute the values of [tex]P[/tex] , [tex]{V_i}[/tex] and [tex]{V_f}[/tex] in above expression .
[tex]\begin{aligned}W&= \left( {1.65 \times {{10}^5}} \right) \times \left( {0.320 - 0.110} \right)\\&= 3.465 \times {10^4}\,{\text{J}}\\&\approx 3.{\text{47}} \times {\text{1}}{{\text{0}}^4}\,{\text{J}}\\\end{aligned}[/tex]
Therefore, the work done by the gas is [tex]\boxed{3.47 \times {{10}^4}\,{\text{J}}}[/tex]
Part (b):
The change in internal energy of the gas is given by the first law of thermodynamics:
[tex]\begin{aligned}\Delta Q = \Delta U + W \hfill\\\Delta U = \Delta Q - W \hfill\\\end{aligned}[/tex]
Here, [tex]\Delta Q[/tex] is the heat added, [tex]\Delta U[/tex] is the change in internal energy of system and [tex]W[/tex] is the work done by the gas.
Substitute the values of [tex]\Delta Q[/tex] and [tex]W[/tex] in above equation.
Part (c):
The ideal gas or the real gas does not affect the process of the expansion of gas because the change in energy of the gas is given by the First law of thermodynamics. The first law of thermodynamics is based on the conservation of energy.
The energy of a system will always remain conserved whether it has the real gas or the ideal gas.
Thus, there is no effect of the gas being ideal gas or real gas on the work done by the gas.
Learn More:
1. In the calorimetry experiment which energy will be calculated during the heat exchange https://brainly.com/question/2566525
2. According to Charles’ law, for a fixed quantity of gas at constant pressure, https://brainly.com/question/7316997
3.What is the kinetic energy of the emitted electrons when cesiumhttps://brainly.com/question/9059731
Answer Details:
Grade: High School
Subject: Physics
Chapter: Law of Thermodynamics
Keywords: Ideal gas, cylinder, first law of thermodynamics, pressure constant, 0.110m^3 to 0.320 m^3, expansion, change in internal energy, work done by gas.
