Question

A sample of helium gas at a pressure of 0.826 atm and a temperature of 203 °C, occupies a volume of 531 mL. If the gas is heated at constant pressure until its volume is 660 mL, the temperature of the gas sample will be
°C.

Answer 1

Given:
Initial pressure (P1) = 0.826 atm
Initial volume (V1) = 531 mL
Initial temperature (T1) = 203 °C = 476 K

Final volume (V2) = 660 mL

Using the combined gas law equation:
(P1V1)/T1 = (P2V2)/T2

Substitute the given values and solve for T2:
(0.826 atm * 531 mL) / 476 K = (0.826 atm * 660 mL) / T2

T2 = (0.826 atm * 660 mL * 476 K) / (0.826 atm * 531 mL)
T2 = 578 K

Therefore, the temperature of the gas sample will be 578 K, which is equivalent to 305 °C.