Question

A sample of gas initially has a volume of 745 mL at 215°C and 3270 mm Hg. What pressure (mm Hg) will the sample have if the volume changes to 3.64 L while the temperature is decreased to 105°C?

• 12400 mm Hg
O 864 mm Hg
O 327 mm Hg
• 518 mm Hg
O 0.0481 mm Hg

Answer 1

Using the combined gas law, after converting temperatures to Kelvin and volume to liters, the pressure of the gas at 3.64 L and 105°C is calculated to be approximately 864 mm Hg.

Step-by-step explanation:

We need to use the combined gas law, which relates pressure, volume, and temperature of a gas. The formula is (P1 × V1) / T1 = (P2 × V2) / T2, where P is pressure, V is volume, and T is temperature in Kelvin. We must first convert all units to the appropriate ones: Celsius to Kelvin by adding 273.15, mL to L by dividing by 1000, and working with a consistent unit for pressure.

First, convert temperatures from °C to K:

Initial temperature: 215°C + 273.15 = 488.15 K

Final temperature: 105°C + 273.15 = 378.15 K

Next, convert the initial volume from mL to L:

Initial volume: 745 mL / 1000 = 0.745 L

Now we can use the combined gas law:

(3270 mm Hg × 0.745 L) / 488.15 K = (P2 × 3.64 L) / 378.15 K

Solving for P2 gives:

P2 = (3270 mm Hg × 0.745 L × 378.15 K) / (488.15 K × 3.64 L)

P2 = ≈864 mm Hg

Therefore, the pressure of the gas at 3.64 L and 105°C is approximately 864 mm Hg.