A cylinder is closed by a piston connected to a spring of constant 2.30 x 103 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C. (a) If the piston has cross-sectional area of 0.0150 m² and negligible mass, how high will it rise when the temperature s raised to 250°C? Note that when the piston rises the volume of the gas increases and the pressure increases because the spring exerts a force on the piston. m (b) What is the pressure of the gas at 250°C? x If you know the displacement of the piston, you can calculate the pressure from the force exerted by the spring, or by considering the effect of the change in volume and temperature using the ideal gas law. Pa

A cylinder is closed by a piston connected to a spring of constant 2.30 x 103 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C. (a) If the piston has cross-sectional area of 0.0150 m² and negligible mass, how high will it rise when the temperature s raised to 250°C? Note that when the piston rises the volume of the gas increases and the pressure increases because the spring exerts a force on the piston. m (b) What is the pressure of the gas at 250°C? x If you know the displacement of the piston, you can calculate the pressure from the force exerted by the spring, or by considering the effect of the change in volume and temperature using the ideal gas law. Pa

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is compressed to 31.0% of its original volume and the temperature is increased to 48.0°C. (a) What is the tire pressure in pascals?(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?
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.
A high-pressure gas cylinder contains 70.0 L of toxic gas at a pressure of 2.10 x 107 N/m2 and a temperature of 14.0°C. Its valve leaks after the cylinder is dropped. The cylinder is cooled to dry Ice temperatures (-78.5°C), to reduce the leak rate and pressure so that it can be safely repaired. (a) What is the final pressure in the tank, assuming a negligible amount of gas leaks while being cooled and that there is no phase change? N/m² (b) What is the final pressure if one-tenth of the gas escapes? N/m² (c) To what temperature must the tank be cooled to reduce the pressure to 1.00 atm (assuming the gas does not change phase and that there is no leakage during cooling)? K Additional Materials
A high-pressure gas cylinder contains 50.13 L of toxic gas at a pressure of 1.40 × 10¹N/m² and a temperature of 25.0°C. Its value leaks after the cylinder is dropped. The cylinder is cooled to dry ice temperature (-78.5°C) to reduce the leak rate and pressure so that it can be safely repaired. (a) What is the final pressure in the tank, assuming a negligible amount of gas leaks while being cooled and that there is no phase change?
A glass container encloses a low pressure of 02 gas at 298 K. The molecular weight of 02 is 32.0 g/mol. (A) Calculate the mean squared velocity, v², of these molecules in units of m²/s². (B) Find the root mean squared speed along the x-axis, Vx. (C) Would Vx be lower, higher, or the same under the same conditions if the gas were He instead of 02? Please explain your answer in 15 words or less.
An 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
(6) There is a cylindrical container with a piston with diameter d: volume to change. Sitting on top of that piston is a m = the diatomic gas is T 4.67cm which allows the 6.62kg block. The temperature of 382K and there are N 1.45 x 1023 molecules. || (a) What is the gauge pressure of the gas inside the container? (b) What is the volume of this container in cm³? (c) What is the mean free path of a molecule in this container? (d) When the block is remo sure matches external air pressure. How much distance will the piston move during this expansion? (e) What is the work done on this gas by the surroundings while the piston moves to its new position? noved, the gas will expand isothermally until the internal pres-
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.