A rigid tank is divided into two equal parts by a partition. Initially, one side of the tank contains 2 kg of water at 250 kPa and 30°C, while the other side is evacuated. The partition is then removed, and the water expands into the entire tank. The water is allowed to exchange heat with its surrounding until the temperature in the tank returns to the initial value of 30°C. Determine (a) the volume of the tank (in m³), (b) the final pressure (in kPa) and (c) the change in internal energy per unit mass (in kJ/kg).
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- Two glass containers, of equal volume each hold a mole of gas. Container 1 is filled with hydrogen gas (molar mass 2 g / mol), and Container 2 holds helium (molar mass 4 g / mol). If the pressure of the gas in Container 1 equals the pressure of the gas in Container 2, which of the following is true? (a) The temperature of the gas in Container 1 is lower than the temperature of the gas in Container 2. (b) The temperature of the gas in Container 1 is greater than the temperature of the gas in Container 2. (c) The value of R for the gas in Container 1 is ½ the value of R for the gas in Container 2. (d) The rms speed of the gas molecules in Container 1 is lower than the rms speed of the gas molecules in Container 2. (e) The rms speed of the gas molecules in Container 1 is greater than the rms speed of the gas molecules in Container 2.Two moles of a helium gas are at a temperature of 260 K. Calculate the average kinetic energy per atom, the root-mean-square (rms) speed of atoms in the gas, and the internal energy of the gas. HINT (a) the average kinetic energy per atom (in J) J (b) the root-mean-square (rms) speed (in m/s) of atoms in the gas m/s (c) the internal energy of the gas (in J) JAn automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is compressed to 27.0% of its original volume and the temperature is increased to 39.0°C. (a) What is the tire pressure in pascals? Pa (b) After the car is driven at high speed, the tire's air temperature rises to 85.0°C and the tire's interior volume increases by 3.00%. What is the new tire pressure (absolute) in pascals? Pa
- 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.In everyday experience, the measures of temperature most often used are Fahrenheit F and Celsius C. Recall that the relationship between them is given by the following formula. F = 1.8C + 32 Physicists and chemists often use the Kelvin temperature scale. You can get kelvins K from degrees Celsius by using the following formula. K = C + 273.15 (a) Calculate that value.K(25) = (b) Find a formula expressing the temperature C in degrees Celsius as a function of the temperature K in kelvins. C = (c) Find a formula expressing the temperature F in degrees Fahrenheit as a function of the temperature K in kelvins. F = (d) What is the temperature in degrees Fahrenheit of an object that is 272 kelvins?On a chilly 10°C day, you quickly take a deep breath—all your lungs can hold, 4.0 L. The air warms to your body temperature of 37°C. If the air starts at a pressure of 1.0 atm, and you hold the volume of your lungs constant (a good approximation) and the number of molecules in your lungs stays constant aswell (also a good approximation), what is the increase in pressure inside your lungs?