Respuesta :
Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J
Explanation :
First we have to calculate the value of [tex]\gamma[/tex].
[tex]\gamma=\frac{C_p}{C_v}[/tex]
As, [tex]C_p=R+C_v[/tex]
So, [tex]\gamma=\frac{R+C_v}{C_v}[/tex]
Given :
[tex]C_v=20.8J/K\\\\R=8.314J/K[/tex]
[tex]\gamma=\frac{8.314+20.8}{20.8}=1.4[/tex]
Now we have to calculate the initial volume of gas.
Formula used :
[tex]P_1V_1=nRT_1[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 4.25 atm
[tex]V_1[/tex] = initial volume of gas = ?
[tex]T_1[/tex] = initial temperature of gas = 300 K
n = number of moles of gas = 1.0 mol
R = gas constant = 0.0821 L.atm/mol.K
[tex](4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)[/tex]
[tex]V_1=5.80L[/tex]
Now we have to calculate the final volume of gas by using reversible adiabatic expansion.
[tex]P_1V_1^{\gamma}=P_2V_2^{\gamma}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 4.25 atm
[tex]P_2[/tex] = final pressure of gas = 2.50 atm
[tex]V_1[/tex] = initial volume of gas = 5.80 L
[tex]V_2[/tex] = final volume of gas = ?
[tex]\gamma[/tex] = 1.4
Now put all the given values in above formula, we get:
[tex](4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}[/tex]
[tex]V_2=8.47L[/tex]
Now we have to calculate the final temperature of gas.
Formula used :
[tex]P_2V_2=nRT_2[/tex]
where,
[tex]P_2[/tex] = final pressure of gas = 2.50 atm
[tex]V_2[/tex] = final volume of gas = 8.47 L
[tex]T_2[/tex] = final temperature of gas = ?
n = number of moles of gas = 1.0 mol
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in above formula, we get:
[tex](2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2[/tex]
[tex]T_2=257.9K\approx 258K[/tex]
Now we have to calculate the work done.
[tex]w=nC_v(T_2-T_1)[/tex]
where,
w = work done = ?
n = number of moles of gas =1.0 mol
[tex]T_1[/tex] = initial temperature of gas = 300 K
[tex]T_2[/tex] = final temperature of gas = 258 K
[tex]C_v=20.8J/K[/tex]
Now put all the given values in above formula, we get:
[tex]w=(1.0mol)\times (20.8J/K)\times (258-300)K[/tex]
[tex]w=-873.6J[/tex]
Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J
Based on the calculations, the final volume of this sample of gas is equal to 8.47 Liters.
Given the following data:
- Number of moles, n = 1.0 mol.
- Heat capacity at constant volume, [tex]C_v[/tex] = 20.8 J/K
- Initial pressure = 4.25 atm.
- Initial temperature = 300 K.
- Final pressure = 2.50 atm.
How to calculate the final volume.
First of all, we would determine the heat capacity ratio for this adiabatic process by using this formula:
[tex]\gamma=\frac{C_p}{C_v} =\frac{R+C_v}{C_v} \\\\\gamma=\frac{8.314 +20.8}{20.8} \\\\\gamma=\frac{29.114}{20.8} \\\\\gamma=1.4[/tex]
Also, we would determine the initial volume of this gas by applying ideal gas equation:
[tex]PV=nRT\\\\V_1=\frac{nRT}{P} \\\\V_1=\frac{1.0 \times 0.0821 \times 300}{4.25}[/tex]
Initial volume = 5.80 Liters.
For the final volume, we would apply the equation of a reversible adiabatic process:
[tex]PV^{\gamma}=k\\\\P_1V_1^{\gamma}=P_2V_2^{\gamma}\\\\(4.25 \times 5.80)^{1.4}=2.50V_2^{1.4}\\\\(24.65 )^{1.4}=2.50V_2^{1.4}\\\\88.83=2.50V_2^{1.4}\\\\V_2^{1.4}=\frac{88.83}{2.50} \\\\V_2^{1.4}=35.532\\\\V_2=\sqrt[1.4]{35.532}[/tex]
Final volume = 8.47 Liters.
How to calculate the final temperature.
[tex]PV=nRT\\\\T_2=\frac{PV}{nR} \\\\T_2=\frac{2.50 \times 8.47 }{1.0 \times 0.0821}\\\\T_2=\frac{21.175 }{ 0.0821}\\\\T_2=257.92[/tex]
Final temperature = 257.92 â 258 Kelvin.
How to calculate the work done.
Mathematically, the work done in an adiabatic process is given by:
[tex]W = nC_v(T_2-T_1)\\\\W=1 \times 20.8(258-300)\\\\W=20.8\times (-42)[/tex]
Work done = -873.6 â 874 Joules.
Read more on adiabatic process here: https://brainly.com/question/3962272
