Question

A sample of argon gas has a volume of 735 mL at a pressure of 1.20 atm and a temperature of 112 ∘C. What is the final volume of the gas, in milliliters, when the pressure and temperature of the gas sample are changed to the following, if the amount of gas does not change? 648 mmHg and 275 K

Answer 1

V2 = 738.57 mL

Step-by-step explanation

Given

Volume 1 = 735 mL

Pressure 1 = 1.20 atm

Temperature 1 = 112 ∘C = 385 K

Pressure 2 = 648 mmHg = 0.853 atm

Temperature 2 = 275 K

Required: Volume 2 in mL

Solution:

Use a combined gas law

`(P_1V_1)/(T_1)\text{ =}(P_2V_2)/(T_2)`

V2 = (P1V1T2)/T1P2

V2 = (1.20 x 735 x 275)/(385 x 0.853)

V2 = 738.57 mL