To solve this problem we will apply the concepts related to the ideal gas equations. Which defines us that the relationship between pressure, temperature and volume in the first state must be equivalent in the second state of matter. In mathematical terms this is
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
If we rearrange the equation to find the Temperature at state 1 we have that
[tex]T_1 = \frac{P_1V_1T_2}{P_2V_2}[/tex]
Replacing our values we have that
[tex]T_1 = \frac{(4.8*5.5*298.15)}{(2.1*9.6)}[/tex]
[tex]T_1 = 390.435K[/tex]
Therefore the temperature is 390.435K
Answer:
117 ∘C
Explanation:
Use the combined gas law.
P1V1/T1 = P2V2/T2
Let the subscript 2 represent the 9.60L of gas at 25.0∘C and the subscript 1 represent the gas at the initial volume of 5.50L.
Remember to covert the temperature from degrees Celsius to Kelvin by adding 273.15.
Therefore, we have that T2=298.15K, P2=2.10atm, V2=9.60L, P1=4.80atm, V1=5.50L, and T1 is unknown.
Rearrange the equation for T1 and substitute in the known values to solve for the initial temperature.
T1T1T1=P1V1T2P2V2=(4.80atm)(5.50L)(298.15K)(2.10atm)(9.60L)=390.434K
Now, convert this temperature from Kelvin to degrees Celsius.
T1=390.434K−273.15 = 117.28∘C
Therefore, after rounding this value to three significant figures, we find that the initial temperature is 117∘C.
