Question

A sample of argon has a volume of 1.2 L at STP. If the temperature is

increased to 22°C and the pressure is lowered to 0.80 atm, what will
the new volume be, in L?​

Answer 1

To find the new volume at the given temperature and pressure, we can use the combined gas law equation. The new volume is 0.876 L.

Step-by-step explanation:

To solve this problem, we can use the combined gas law equation, which states that P1V1/T1 = P2V2/T2. Given that the initial volume (V1) is 1.2 L, the initial pressure (P1) is 1 atm (STP), the initial temperature (T1) is 273 K, and the new pressure (P2) is 0.80 atm, we need to solve for the new volume (V2) at the new temperature (T2). We can convert the temperature from Celsius to Kelvin by adding 273 to 22°C, giving us 295 K (T2).

Using the formula, we have:

1(1.2)/273 = 0.80(V2)/295

Simplifying the equation, we find that V2 = (0.80)(1.2)(295)/273 = 0.876 L.

Answer 2

the new volume would be 1.62 L

Step-by-step explanation:

stp is 0 degrees C or 273.15 K and 1atm for the pressure

22 degrees C is 295.15 K

use the combined gas law formula and solve for v2

v2=p1(v1)(t2)/p2(t1)

v2=1atm(1.2L)(295.15)/0.8atm(273.15K)

v2=1.62