Respuesta :
Explanation:
It is given that initially pressure of ideal gas is 4.00 atm and its temperature is 350 K. Let us assume that the final pressure is [tex]P_{2}[/tex] and final temperature is [tex]T_{2}[/tex].
(a) We know that for a monoatomic gas, value of [tex]\gamma[/tex] is \frac{5}{3}[/tex].
And, in case of adiabatic process,
[tex]PV^{\gamma}[/tex] = constant
also, PV = nRT
So, here [tex]T_{1}[/tex] = 350 K, [tex]V_{1} = V[/tex], and [tex]V_{2} = 1.5 V[/tex]
Hence, [tex]\frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma -1}[/tex]
[tex]\frac{T_{2}}{350 K} = (\frac{V}{1.5V})^{\frac{5}{3} -1}[/tex]
[tex]T_{2}[/tex] = 267 K
Also, [tex]P_{1}[/tex] = 4.0 atm, [tex]V_{1} = V[/tex], and [tex]V_{2} = 1.5 V[/tex]
[tex]\frac{P_{2}}{P_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma}[/tex]
[tex]\frac{P_{2}}{4.0 atm} = (\frac{V}{1.5V})^{\frac{5}{3}}[/tex]
[tex]P_{2}[/tex] = 2.04 atm
Hence, for monoatomic gas final pressure is 2.04 atm and final temperature is 267 K.
(b) For diatomic gas, value of [tex]\gamma[/tex] is \frac{7}{5}[/tex].
As, [tex]PV^{\gamma}[/tex] = constant
also, PV = nRT
[tex]T_{1}[/tex] = 350 K, [tex]V_{1} = V[/tex], and [tex]V_{2} = 1.5 V[/tex]
[tex]\frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma -1}[/tex]
[tex]\frac{T_{2}}{350 K} = (\frac{V}{1.5V})^{\frac{7}{5} -1}[/tex]
[tex]T_{2}[/tex] = 289 K
And, [tex]P_{1}[/tex] = 4.0 atm, [tex]V_{1} = V[/tex], and [tex]V_{2} = 1.5 V[/tex]
[tex]\frac{P_{2}}{P_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma}[/tex]
[tex]\frac{P_{2}}{4.0 atm} = (\frac{V}{1.5V})^{\frac{7}{5}}[/tex]
[tex]P_{2}[/tex] = 2.27 atm
Hence, for diatomic gas final pressure is 2.27 atm and final temperature is 289 K.
