174k views
3 votes
A sample of methane, CH₄, occupies a volume of 170.0 ml at 25°c and exerts a pressure of 965.0 mmg. if the volume of the gas is allowed to expand to 720.0 ml at 298 k, what will be the pressure of the gas?

a. 4090 mmHg
b. 2720 mmHg
c. 92.8 mmHg
d. 0.008823 mmHg
e. 550.0 mmHg

User Duy
by
7.9k points

1 Answer

1 vote

Final answer:

The pressure of the gas will be approximately 92.8 mmHg.

Step-by-step explanation:

To find the pressure of the gas, we can use the combined gas law equation: P1V1 / T1 = P2V2 / T2, where P1 = 965.0 mmHg (initial pressure), V1 = 170.0 ml (initial volume), T1 = 25°C (initial temperature), V2 = 720.0 ml (final volume), and T2 = 298 K (final temperature).

Plugging in the values, we get:

(965.0 mmHg)(170.0 ml) / (25 + 273) K = P2(720.0 ml) / 298 K

P2 = [(965.0 mmHg)(170.0 ml) / (25 + 273) K] / (720.0 ml / 298 K) ≈ 92.8 mmHg

Therefore, the pressure of the gas will be approximately 92.8 mmHg.

The pressure of a gas when the volume is allowed to expand can be determined using the Combined Gas Law, which is a combination of Boyle's Law, Charles's Law, and Gay-Lussac's Law.

For a sample of methane (CH4) that occupies a volume of 170.0 ml at 25°C (which is 298 K) and exerts a pressure of 965.0 mmHg, to find the new pressure when the volume is expanded to 720.0 ml at the same temperature of 298 K, the Combined Gas Law formula is used:

P1V1/T1 = P2V2/T2

User Simon Scarfe
by
9.0k points