A sample of argon has a volume of 735 mL at a pressure of 1.20 atm and a temperature of 385 K. What is the volume of the gas in milliliters when the pressure and temperature of the gas sample are changed to 15.4 atm and 2580C? This problem involves changes in pressure, temperature, and volume.

(P1V1/T1) = (P2V2/T2)

Remember BOTH temperatures must be in kelvin

To find the volume of the gas when the pressure and temperature are changed, we can use the combined gas law formula. The combined gas law relates the initial and final values of pressure, temperature, and volume of a gas sample.

The formula for the combined gas law is:
(P1 * V1) / (T1) = (P2 * V2) / T2

Where:
P1 and P2 are the initial and final pressures, respectively.
V1 and V2 are the initial and final volumes, respectively.
T1 and T2 are the initial and final temperatures, respectively.

We are given:
P1 = 1.20 atm
V1 = 735 mL
T1 = 385 K

We need to find V2 when:
P2 = 15.4 atm
T2 = 2580°C

First, we need to convert T2 from Celsius to Kelvin:
T2 = 2580°C + 273.15 = 2853.15 K

Now, we can plug the values into the combined gas law formula and solve for V2:

(1.20 atm * 735 mL) / 385 K = (15.4 atm * V2) / 2853.15 K

To isolate V2, we can cross-multiply and then divide:

(1.20 atm * 735 mL * 2853.15 K) = (15.4 atm * V2 * 385 K)

V2 = (1.20 atm * 735 mL * 2853.15 K) / (15.4 atm * 385 K)

V2 ≈ 1235.47 mL

Therefore, the volume of the argon gas is approximately 1235.47 mL when the pressure is changed to 15.4 atm and the temperature is changed to 2580°C.