A 3.00 L tank contains air at 3.00 atm and 20.0C. The tank is sealed and cooled until the pressure is 1.00 atm. (a) What is the temperature then in degrees Celsius? Assume that the volume of the tank is constant. (b) If the temperature is kept at the value found in part (a) and the gas is compressed, what is the volume when the pressure again becomes 3.00 atm?

A 3.00 L tank contains air at 3.00 atm and 20.0C. The tank is sealed and cooled until the pressure is 1.00 atm. (a) What is the temperature then in degrees Celsius? Assume that the volume of the tank is constant. (b) If the temperature is kept at the value found in part (a) and the gas is compressed, what is the volume when the pressure again becomes 3.00 atm?

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(a) A tank contains one mole of helium gas at a pressure of 6.35 atm and a temperature of 22.0°C. The tank (which has a fixed volume) is heated until the pressure inside triples. What is the final temperature of the gas? °C (b) A cylinder with a moveable piston contains one mole of helium, again at a pressure of 6.35 atm and a temperature of 22.0°C. Now, the cylinder is heated so that both the pressure inside and the volume of the cylinder double. What is the final temperature of the gas? °C
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Step 1 (a) The ideal-gas equation describes each condition of the air in the tire. We will set up ratios in order to describe these changes. Dividing the equations PV = nRT in the initial and final states, we have PfVf Tf P¡Vi Ti Solving for the final pressure in the tire, gives Pf = = = which gives I Tf P(V)() P₁ Vf T₁ = (1.013 x 105 Pa) x 105 Pa. Note that the volume V¡ cancels out of the equation. = Vf Po = Pr (V ( ) ( ) Pd Pf Vi V₁ Step 2 (b) After the car is driven, the temperature and volume of air in the tire have changed. Let Td be the temperature and Vd be the volume of air in the tire. We have PdVd Td PfVf Tf' x 105 Pa 273 + 273 + Vf Vf Again we see that the volume Vf cancels out of the equation. °C K °℃ K = x 105 Pa.
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