Final answer:
To find the initial temperature of helium gas changing from 1.56 atm to 0.975 atm at 27.0°C, we use the combined gas law to calculate that the initial temperature was 207.09°C.
Step-by-step explanation:
The student is asking about a problem related to the initial temperature of a helium gas sample undergoing a pressure change while maintaining a constant volume, which is a characteristic of Ideal Gas Law problems.
To find the initial temperature of the helium (T1), we can use the combined gas law formula which is derived from the Ideal Gas Law: P1V1/T1 = P2V2/T2.
Since the volume is constant and cancels out, we are left with P1/T1 = P2/T2.
Rearranging for T1 gives us T1 = (P1/P2) × T2.
Plugging in the values, P1 = 1.56 atm, P2 = 0.975 atm, and T2 = 27.0 °C + 273.15 (to convert to Kelvin), we can solve for T1.
To calculate:
T1 = (1.56 atm / 0.975 atm) × (27.0 °C + 273.15 K)
T1 = (1.60) × (300.15 K)
T1 = 480.24 K
Then we convert T1 back to °C:
T1 = 480.24 K - 273.15
= 207.09°C
Thus, the initial temperature of the helium gas was 207.09°C.