Answer:
The final pressure of the gas after the expansion is 1.37 x 10³ Pa.
Explanation:
Given;
number of gas moles, n = 5
initial pressure of the ideal gas, P₁ = 2500 Pa
initial volume of the ideal gas, V₁ = 2.1 m³
work done by the gas, W = 1680 J
For adiabatic process, we will have;
[tex]P_1V_1^r = P_2V_2^r\\\\P_2 = \frac{P_1V_1^r}{V_2^r}[/tex]
For monatomic gas, r = 1.67
[tex]P_2 = \frac{2500*(2.1)^{1.67}}{V_2^{1.67}}\\\\P_2 = \frac{8630.704}{V_2^{1.67}} -----equation \ (1)[/tex]
For work done by the gas;
[tex]W = \frac{1}{r-1}(P_1V_1 -P_2V_2)\\\\W = \frac{P_1V_1 -P_2V_2}{r-1} \\\\1680 = \frac{2500*2.1-P_2V_2}{1.67-1} \\\\1680 = \frac{5250-P_2V_2}{0.67} \\\\5250-P_2V_2 = 1125.6\\\\P_2V_2 = 5250 - 1125.6\\\\P_2V_2 = 4124.4\\\\P_2 = \frac{4124.4}{V_2} ------equation \ (2)[/tex]
Solve equation (1) and (2) together
[tex]\frac{4124.4}{V_2} = \frac{8630.704}{V_2^{1.67}}\\\\\frac{V_2^{1.67}}{V_2} = \frac{8630.704}{4124.4}\\\\\frac{V_2^{1.67}}{V_2} =2.0926\\\\V_2^{1.67-1} =2.0926\\\\V_2^{0.67} =2.0926\\\\V_2 = (2.0926)^{\frac{1}{0.67}} \\\\V_2 = 3.01 \ m^3[/tex]
Now, determine the final pressure of the gas using equation (2)
[tex]P_2 = \frac{4124.4}{V_2} \\\\P_2 = \frac{4124.4}{3.01}\\\\P_2 = 1370.23 \ Pa\\\\P_2 = 1.37*10^{3} \ Pa[/tex]
Therefore, the final pressure of the gas after the expansion is 1.37 x 10³ Pa.
