Question

A)High-pressure liquid chromatography (HPLC) is a method used in chemistry and biochemistry to purify chemical substances. The pressures used in this procedure range from around 500 kilopascals (500,000 Pa) to about 60,000 kPa (60,000,000 Pa). It is often convenient to know the pressure in torr. If an HPLC procedure is running at a pressure of 2.04×108 Pa , what is its running pressure in torr?
B)A sample of gas in a balloon has an initial temperature of 37 ∘C and a volume of 1.12×103 L . If the temperature changes to 58 ∘C , and there is no change of pressure or amount of gas, what is the new volume, V2, of the gas?

C)A very flexible helium-filled balloon is released from the ground into the air at 20. ∘C. The initial volume of the balloon is 5.00 L, and the pressure is 760. mmHg. The balloon ascends to an altitude of 20 km, where the pressure is 76.0 mmHg and the temperature is −50. ∘C. What is the new volume, V2, of the balloon in liters, assuming it doesn't break or leak?

D )Consider 4.60 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.40 L and the temperature is increased to 30. ∘C , what is the new pressure, P2, inside the container? Assume no change in the amount of gas inside the cylinder.

Answer 1

Answer:

Question 1. To convert  2.04x10⁸Pa to Torr?

From google, 1 Torr = 133.32Pa

Therefore, 2.04x10⁸Pa = 2.04x10⁸Pa x 1Torr/133.32Pa

                             Pressure = 1.53x10⁶Torr

Question 2. Given state 1 : V=1.12x10³L=1.12m³ T=37C= 310K

                                state 2 : T=58C = 331K, V=?

But we are told Pressure is constant, we will use Charle's law equation to get V

V₁T₂=V₂T₁,  

V₂ = V₁T₂/T₁  === 1.12 x 331/310 = 1.195m³

Question 3:

Given state 1 : V=5.00L=0.005m³, T=20C= 293K, P = 760mmHg

                                state 2 : V=? T= -50C = 223K, P = 76mmHg

But we are told Pressure is constant, we will use general gas law equation to get V

P₁V₁T₂=P₂V₂T₁,  

V₂ = P₁V₁T₂/P₂T₁  === 760mmHg x 0.005m³ x 223K/(293K x 76mmHg) =                  0.038m³

Question 4:

Given state 1 : V=4.60L=0.0046m³, T=20C= 293K, P = 365mmHg

                                state 2 : V=2.4L = 0.0024m³, T= 30C = 303K, P = ?

But we are told Pressure is constant, we will use general gas law equation to get V

P₁V₁T₂=P₂V₂T₁,  

P₂ = P₁V₁T₂/V₂T₁  

===  365mmHg x 0.0046m³ x 303K/(293K x 0.0024m³)                                          =         723.5mmHg