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Unformatted text preview: Oregon State University Physics 202 , Winter 2010 HW67 6 (due Feb. 22 at 5:00 p.m.) 7 Page 1 Oregon State University Physics 202 Winter Term, 2010 HW67 1. A spherical balloon is made from a rubber material. The mass of the rubber is 4.00 kg. The thick ness of the rubber is negligible compared to the 2.00m radius of the balloon. The balloon is F lled with helium (He). The temperature of the helium and all the surrounding air is 25°C, and the balloon just ﬂ oats in the air (neither rising nor sinking). If the density of the surrounding air is 1.20 kg/m 3 , what is the absolute pressure of the helium? The balloon “just ﬂ oats.” Thus: ρ balloon = ρ air By def nition: ρ balloon = m balloon / V balloon More specif cally: ρ balloon = ( m rubber + m He )/[(4/3) π r balloon 3 ] The air has the same density: ρ air = ( m rubber + m He )/[(4/3) π r balloon 3 ] Solve this For m He .… ρ air [(4/3) π r balloon 3 ] = ( m rubber + m He ) ρ air [(4/3) π r balloon 3 ] – m rubber = m He But by def nition: m He = n He m mole.He Or: m He / m mole.He = n He Substitute the above result: { ρ air [(4/3) π r balloon 3 ] – m rubber }/ m mole.He = n He Apply the ideal gas law to the He: P He V He = n He RT He Or: P He = n He RT He / V He Substitute From above: P He = ({ ρ air [(4/3) π r balloon 3 ] – m rubber }/ m mole.He ){ RT He /[(4/3) π r balloon 3 ]} Plug in the numbers: P He = ({(1.20)[(4/3) π (2.00) 3 ] – 4.00}/.004){(8.31)(298)/[(4/3) π (2.00) 3 ]} P He = 6.69 x 10 5 Pa Oregon State University Physics 202 , Winter 2010 HW67 6 (due Feb. 22 at 5:00 p.m.) 7 Page 2 2. A cylinder containing 6.95 moles of helium gas is initially Before After as shown in the F rst drawing here. The cylinder cover is free to slide up or down the cylinder, but it is at rest. The space above the cylinder cover is open to the atmosphere. Then a block is placed on the cylinder cover and the cover moves slowly into the other position shown. When it has arrived at that position (and is again at rest), the temper ature of the helium has increased by 10.0°C. Using the above information and the following data, determine the mass of the block. d = 0.623 m; d d h 1 = 5.17 m; h 2 = 4.83 m The mass of the cylinder cover is 5.24 kg For an ideal gas: PV = nRT ... and the number of moles ( n ) stays constant. Therefore: P i V i / T i = P f V f f / f / f T f Solving for P f : P f = ( f f T f / f / f T i )( V i / V f ) f P i Substituting: P f = [( f f T i + 10)/ T i )]( h 1 / h 2 ) P i = [ + (10/ T i )]( h 1 / h 2 ) P i where: A = π d 2 d /4 = π (0.623) 2 /4 = π (0.623) 2 /4 = 0.304836 m 2 and: P i = P atm + m cyl g / A = 1.01 x 10 5 + (5.24)(9.80)/0.304836 = 101168 Pa and: T i = P i V i /( nR ) = P i ( h 1 )( A )/( nR ) = 101168(5.17)(0.304836)/[(6.95)(8.31)] = 2760.68K and: P f = f f P atm + ( m cyl + m block ) g / A Put it all together: P atm + ( m cyl + m block ) g / A = [1 + (10/ T i )]( h 1 / h 2 )(101168) Now solve for m block : Subtract P atm from both sides:...
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This note was uploaded on 03/27/2010 for the course PH 202 taught by Professor Staff during the Spring '08 term at Oregon State.
 Spring '08
 Staff
 Physics, Mass

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