Answer:
temperature on left side is 1.48 times the temperature on right
Explanation:
GIVEN DATA:
![\gamma = 5/3](https://tex.z-dn.net/?f=%5Cgamma%20%3D%205%2F3)
T1 = 525 K
T2 = 275 K
We know that
![P_1 = \frac{nRT_1}{v}](https://tex.z-dn.net/?f=P_1%20%3D%20%5Cfrac%7BnRT_1%7D%7Bv%7D)
![P_2 = \frac{nrT_2}{v}](https://tex.z-dn.net/?f=P_2%20%3D%20%5Cfrac%7BnrT_2%7D%7Bv%7D)
n and v remain same at both side. so we have
![\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1%7D%7BP_2%7D%20%3D%20%5Cfrac%7BT_1%7D%7BT_2%7D%20%3D%20%5Cfrac%7B525%7D%7B275%7D%20%3D%20%5Cfrac%7B21%7D%7B11%7D)
..............1
let final pressure is P and temp ![T_1 {f} and T_2 {f}](https://tex.z-dn.net/?f=T_1%20%7Bf%7D%20and%20T_2%20%7Bf%7D)
![P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}](https://tex.z-dn.net/?f=P_1%5E%7B1-%5Cgamma%7D%20T_1%5E%7B%5Cgamma%7D%20%3D%20P%5E%7B1%20-%20%5Cgamma%7DT_1%20%7Bf%7D%5E%7B%5Cgamma%7D)
..................2
similarly
.............3
divide 2 equation by 3rd equation
![\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}](https://tex.z-dn.net/?f=%5Cfrac%7B21%7D%7B11%7D%5E%7B-2%2F3%7D%20%5Cfrac%7B21%7D%7B11%7D%5E%7B5%2F3%7D%20%3D%20%5B%5Cfrac%7BT_1%20%7Bf%7D%7D%7BT_2%20%7Bf%7D%7D%5D%5E%7B5%2F3%7D)
![T_1 {f} = 1.48 T_2 {f}](https://tex.z-dn.net/?f=T_1%20%7Bf%7D%20%3D%201.48%20T_2%20%7Bf%7D)
thus, temperature on left side is 1.48 times the temperature on right