Question

A compressor takes 0.50 m3 of a gas at 33°C and 760 mm Hg and compresses it to 0.10 m3, cooling it to -55°C at the same time. What is the pressure of the gas at these new conditions?

3.7 x 10-4 mm Hg
68 mm Hg
2,700 mm Hg
1.0 mm Hg

Answer 1

The pressure of the gas at these new conditions is 2700 mmHg.

Step-by-step explanation:

Data:

initial pressure, P1 = 760 mmHg

initial volume, V1 = 0.5 m^3

initial temperature, T1 = 33 + 273 = 306 K

final pressure, P2 = ? mmHg

final volume, V2 = 0.1 m^3

final temperature, T2 = -55 + 273 = 218 K

Using the Combined gas law we get:

P1*V1/T1 = P2*V2/T2

Solving for P2 and replacing with data (dimensions are omitted):

P2 = P1*V1*T2/(T1*V2)

P2 = 760*0.5*218/(306*0.1)

P2 ≈ 2700 mmHg

Answer 2

The pressure of the gas is 2700 mmHg.

To solve this problem, we can use the Combined Gas Laws:

(p_1V_1)/T_1 = (p_2V_2)/T_2

p_2 = p_1 × V_1/V_2 × T_2/T_1

T_2 = (-55+273.15) K = 218.15 K; T_1 = (33+273.15) K = 306.15 K

p2 = 760 mmHg × (0.50 m^3/0.10 m^3) × (218.15 K/306.15 K) = 2700 mmHg