Question

It is calculated that an experiment will yield 8.36 L of methane gas at STP. If the actual temperature is 42° C and the actual pressure is 622 mmHg, what volume of methane would result? (1 bar = 0.987 atm and 1 atm = 760 mmHg)

Answer 1

Applying the combined gas law, we find that the actual volume of methane gas at 42°C and 622 mmHg is 11.83 liters.

Step-by-step explanation:

To calculate the volume of methane at non-STP conditions, we can apply the combined gas law, which is derived from Boyle's Law, Charles's Law, and Avogadro's Law. The formula for the combined gas law is: P1V1/T1 = P2V2/T2, where P is the pressure, V is the volume, and T is the temperature in Kelvins. Given the volume (V1) is 8.36 L at STP (P1 = 1 atm and T1 = 273.15 K), we want to find the volume (V2) at the actual conditions (P2 = 622 mmHg / 760 mmHg per atm, T2 = 42°C + 273.15 = 315.15 K).

First, convert the actual pressure from mmHg to atm: P2 = 622 mmHg / 760 mmHg/atm = 0.8184 atm. Now, plug in the values into the combined gas law and solve for V2:

1. P1V1/T1 = P2V2/T2
2. (1 atm × 8.36 L) / 273.15 K = (0.8184 atm × V2) / 315.15 K
3. ((1 × 8.36) / 273.15) × 315.15 = 0.8184 × V2
4. 9.677 L = 0.8184 × V2
5. V2 = 9.677 L / 0.8184 atm
6. V2 = 11.83 L

Thus, the volume of methane gas at the actual conditions of 42°C and 622 mmHg is 11.83 liters.