The new volume of the sample of argon, in L, will be 1.43 L.
At STP (Standard Temperature and Pressure), the temperature is 0 °C (273 K) and the pressure is 1 atm. Therefore, the initial conditions can be expressed as:
T1 = 273 K
P1 = 1 atm
V1 = 1.20 L
To find the new volume, we can use the combined gas law:
(P1V1) / T1 = (P2V2) / T2
We can rearrange this equation to solve for V2:
V2 = (P1V1T2) / (P2T1)
Substituting the given values, we get:
V2 = (1 atm x 1.20 L x 301 K) / (0.800 atm x 273 K)
V2 = 1.43 L
At a temperature of 28.0 °C and a pressure of 0.800 atm, the new volume of the argon sample is 1.43 L.
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