Answer:
The final pressure of the gas is 9.94 atm.
Explanation:
Given that,
Weight of argon = 0.16 mol
Initial volume = 70 cm³
Angle = 30°C
Final volume = 400 cm³
We need to calculate the initial pressure of gas
Using equation of ideal gas
[tex]PV=nRT[/tex]
[tex]P_{i}=\dfrac{nRT}{V}[/tex]
Where, P = pressure
R = gas constant
T = temperature
Put the value in the equation
[tex]P_{i}=\dfrac{0.16\times8.314\times(30+273)}{70\times10^{-6}}[/tex]
[tex]P_{i}=5.75\times10^{6}\ Pa[/tex]
[tex]P_{i}=56.827\ atm[/tex]
We need to calculate the final temperature
Using relation pressure and volume
[tex]P_{2}=\dfrac{P_{1}V_{1}}{V_{2}}[/tex]
[tex]P_{2}=\dfrac{56.827\times70}{400}[/tex]
[tex]P_{2}=9.94\ atm[/tex]
Hence, The final pressure of the gas is 9.94 atm.
The final pressure of the gas given that it undergoes an isothermal expansion is 9.95 atm
The initial pressure of gas can be obtained by using the ideal gas equation as illustrated below:
P = nRT / V
P = (0.16 × 0.0821 × 303) / 0.07
P = 56.86 atm
P₁V₁ = P₂V₂
56.86 × 70 = P₂ × 400
Divide both side by 400
P₂ = (56.86 × 70) / 400
P₂ = 9.95 atm
Learn more about gas laws:
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