To determine the final pressure of helium gas when the volume is reduced from 250 ml to 100 ml at a temperature of 45 °C, we can use the combined gas law equation, which relates the initial and final states of a gas sample:
(P1 * V1) / (T1 * n1) = (P2 * V2) / (T2 * n2)
Where: P1 and P2 are the initial and final pressures, respectively. V1 and V2 are the initial and final volumes, respectively. T1 and T2 are the initial and final temperatures in Kelvin, respectively. n1 and n2 represent the number of moles of gas and can be assumed to be constant.
Given: V1 = 250 ml = 250 cm³ V2 = 100 ml = 100 cm³ T1 = 0 °C = 273.15 K (adding 273.15 to convert from Celsius to Kelvin) T2 = 45 °C = 45 + 273.15 = 318.15 K
Assuming the number of moles of gas remains constant, we can simplify the equation:
(P1 * V1) / T1 = (P2 * V2) / T2
Now, we can substitute the known values into the equation:
(1 atm * 250 cm³) / 273.15 K = (P2 * 100 cm³) / 318.15 K
Simplifying further:
P2 = (1 atm * 100 cm³ * 273.15 K) / (250 cm³ * 318.15 K)
P2 ≈ 0.270 atm
Therefore, the final pressure of helium gas, when the volume is reduced to 100 ml at a temperature of 45 °C, is approximately 0.270 atm.