A compressor takes 0.50 m3 of a gas at 33°C and 760 mmHg 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 mmHg 68 mmHg 2,700 mmHg 1.0 mmHg

Respuesta :

Answer:

2,700 mmHg.

Explanation:

  • We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If n is constant, and have two different values of V, P and T:

P₁V₁T₂ = P₂V₂T₁

P₁ = 760 mmHg, V₁ = 0.50 m³, T₁ = 33°C + 273 = 306.0 K.

P₂ = ??? mmHg, V₂ = 0.10 m³, T₂ = - 55°C + 273 = 218.0 K.

∴ P₂ = P₁V₁T₂/V₂T₁ = (760 mmHg)(0.50 m³)(218.0 K)/(0.10 m³)(306.0 K) = 2707.0 mmHg ≅  2,700 mmHg.

Answer:

2,700 mmHg.

Explanation: