The new volume of the helium sample would be 2.4 L.
Volume of a gasAccording to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in kelvins.
At STP (standard temperature and pressure), which is defined as 0°C (273.15 K) and 101.325 kPa, the volume of 2.0 liters of helium gas contains one mole of helium atoms.
To find the new volume of the helium sample when the temperature is increased to 27°C (300.15 K) and the pressure is decreased to 80 kPa, we can use the following equation:
(P1V1)/T1 = (P2V2)/T2
where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.
Plugging in the values, we get:
(101.325 kPa)(2.0 L)/(273.15 K) = (80 kPa)(V2)/(300.15 K)
Solving for V2, we get:
V2 = (101.325 kPa)(2.0 L)/(273.15 K) * (300.15 K)/(80 kPa) = 2.36 L
Therefore, the new volume of the helium sample is approximately 2.4 L (rounded to the nearest tenth).
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