A sample of argon has a volume of 1.20 L at STP. If the temperature is increased to 28.0 °C and the pressure is lowered to 0.800 atm, what will the new volume be, in L?

Answers

Answer 1

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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