Acompressor 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 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

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

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

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.

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