Answer:
When the volume increased from 2.00 to 5.25L the new temperature is 808.9 K ( =535.75 °C)
Explanation:
Step 1: Data given
The initial volume of the sample = 2.00 L
The initial temperature = 35 °C = 308 K
The increased volume = 5.25 L
Pressure = constant
Step 2: Calculate the new temperature
V1/T1 = V2/T2
⇒ with V1 = the initial volume = 2.00 L
⇒ with T1 = the initial volume = 308 K
⇒ with V2 = the new volume = 5.25 L
⇒ with T2 = the new temperature
2.00 / 308 = 5.25 / T2
0.00649 = 5.25/T2
T2 = 5.25/ 0.00649
T2 = 808.9 K
When the volume increased from 2.00 to 5.25L the new temperature is 808.9 K ( =535.75 °C)
