The volume of a sample of hydrogen gas is 246 mL at STP (standard temperature and pressure). what is the temperature of the gas, when the pressure is then increased 1.20 atm and the volume is decreased to 225 mL?
(1.20)(225)/T = (1)(246)/273
To solve this problem, you can use the combined gas law, which relates the pressure, volume, and temperature of a gas. The combined gas law formula is given by:
(P1 * V1) / T1 = (P2 * V2) / T2,
where P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, and T1 and T2 are the initial and final temperatures.
So, let's plug in the given values into the formula and solve for T2:
Initial pressure, P1 = 1 atm (since it is at STP)
Initial volume, V1 = 246 mL
Final pressure, P2 = 1.20 atm
Final volume, V2 = 225 mL
Substituting these values into the equation:
(1 atm * 246 mL) / T1 = (1.20 atm * 225 mL) / T2.
Now, let's solve for T2 by cross-multiplying and rearranging the equation:
T2 = (T1 * (1.20 atm * 225 mL)) / (1 atm * 246 mL).
Since we know the temperature at STP is 273.15 K, we can substitute T1 with this value:
T2 = (273.15 K * (1.20 atm * 225 mL)) / (1 atm * 246 mL).
Now, let's calculate T2:
T2 = (273.15 K * 270 atm·mL) / (246 mL).
Canceling out the units:
T2 ≈ 298.6 K.
Therefore, the temperature of the gas, when the pressure is increased to 1.20 atm and the volume is decreased to 225 mL, is approximately 298.6 K.