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
.
![\gamma=\frac{C_p}{C_v}](https://tex.z-dn.net/?f=%5Cgamma%3D%5Cfrac%7BC_p%7D%7BC_v%7D)
As, ![C_p=R+C_v](https://tex.z-dn.net/?f=C_p%3DR%2BC_v)
So, ![\gamma=\frac{R+C_v}{C_v}](https://tex.z-dn.net/?f=%5Cgamma%3D%5Cfrac%7BR%2BC_v%7D%7BC_v%7D)
Given :
![C_v=20.8J/K\\\\R=8.314J/K](https://tex.z-dn.net/?f=C_v%3D20.8J%2FK%5C%5C%5C%5CR%3D8.314J%2FK)
![\gamma=\frac{8.314+20.8}{20.8}=1.4](https://tex.z-dn.net/?f=%5Cgamma%3D%5Cfrac%7B8.314%2B20.8%7D%7B20.8%7D%3D1.4)
Now we have to calculate the initial volume of gas.
Formula used :
![P_1V_1=nRT_1](https://tex.z-dn.net/?f=P_1V_1%3DnRT_1)
where,
= initial pressure of gas = 4.25 atm
= initial volume of gas = ?
= initial temperature of gas = 300 K
n = number of moles of gas = 1.0 mol
R = gas constant = 0.0821 L.atm/mol.K
![(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)](https://tex.z-dn.net/?f=%284.25atm%29%5Ctimes%20V_1%3D%281.0mol%29%5Ctimes%20%280.0821L.atm%2Fmol.K%29%5Ctimes%20%28300K%29)
![V_1=5.80L](https://tex.z-dn.net/?f=V_1%3D5.80L)
Now we have to calculate the final volume of gas by using reversible adiabatic expansion.
![P_1V_1^{\gamma}=P_2V_2^{\gamma}](https://tex.z-dn.net/?f=P_1V_1%5E%7B%5Cgamma%7D%3DP_2V_2%5E%7B%5Cgamma%7D)
where,
= initial pressure of gas = 4.25 atm
= final pressure of gas = 2.50 atm
= initial volume of gas = 5.80 L
= final volume of gas = ?
= 1.4
Now put all the given values in above formula, we get:
![(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}](https://tex.z-dn.net/?f=%284.25atm%29%5Ctimes%20%285.80L%29%5E%7B1.4%7D%3D%282.50atm%29%5Ctimes%20V_2%5E%7B1.4%7D)
![V_2=8.47L](https://tex.z-dn.net/?f=V_2%3D8.47L)
Now we have to calculate the final temperature of gas.
Formula used :
![P_2V_2=nRT_2](https://tex.z-dn.net/?f=P_2V_2%3DnRT_2)
where,
= final pressure of gas = 2.50 atm
= final volume of gas = 8.47 L
= 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:
![(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2](https://tex.z-dn.net/?f=%282.50atm%29%5Ctimes%20%288.47L%29%3D%281.0mol%29%5Ctimes%20%280.0821L.atm%2Fmol.K%29%5Ctimes%20T_2)
![T_2=257.9K\approx 258K](https://tex.z-dn.net/?f=T_2%3D257.9K%5Capprox%20258K)
Now we have to calculate the work done.
![w=nC_v(T_2-T_1)](https://tex.z-dn.net/?f=w%3DnC_v%28T_2-T_1%29)
where,
w = work done = ?
n = number of moles of gas =1.0 mol
= initial temperature of gas = 300 K
= final temperature of gas = 258 K
![C_v=20.8J/K](https://tex.z-dn.net/?f=C_v%3D20.8J%2FK)
Now put all the given values in above formula, we get:
![w=(1.0mol)\times (20.8J/K)\times (258-300)K](https://tex.z-dn.net/?f=w%3D%281.0mol%29%5Ctimes%20%2820.8J%2FK%29%5Ctimes%20%28258-300%29K)
![w=-873.6J](https://tex.z-dn.net/?f=w%3D-873.6J)
Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J