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

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

Explanation:

dа ta:

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

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