Question

please help!

erniperture cr VOILITE Analyze (A) Find the pressure in the tire at -15 C Rewrite the ideal gas equation PV nRT so that the two variables that change are both on the left-hand side: P nR T V Since all the quantities on the right-hand side of the equation remain constant, the initial value of the left-hand side of the equation must equal the final value: P P T T or P P T A tire gauge measures pressure in excess of atmospheric pressure, while P. and P, used in the ideal gas law are absolute pressures. So, adding atmospheric pressure to the m psi 2.206 x 10 N/m2, gives the absolute pressure: 10 N/m2 P (1.013 x 10 N/m2) + ( 2.206 x 105 N/m2) 3219 C 258K leads to: Substituting in the two temperatures of T, and T,-15 258 K P P 0.8805 x x 10 N/m2 293 K To find the pressure a tire gauge would measure we need to subtract atmospheric pressure from the absolute value just calculated. Using the conversion factor 1 psi 6895 M is: P P-1.013 x 10 N/m6895 N/m 1 psi psi (B) What is the pressure in the pump when you push the piston down as described? The initial volume V, inside the pump is reduced to a final value of V, 0.90 V, thereby ch Using the ideal gas law to relate the initial pressure and volume to the final pressure and volume we find PV-nRT PN Set V 0.90 V, and obtain: 1.013 x 105 N/m2 0.90 P 105 N/m2 0.90 Finalize Are the results of the problem consistent with your experience in observing the change in tire pressure when the weather turns colder?

Answer 1

A)

P_f = [(258K)/(293K)]P_{i =} [(258K)/(293K)](3.219*10⁵ N/m²) = 2.83*10⁵ N/m²

P_gauge = [P_f -1.013*10⁵ N/m²] = [2.83*10⁵ N/m²-1.013*10⁵ N/m²]*(1psi/6895N/m²)= (1.817*10⁵ N/m²)/( 6895N/m²) = 26.35 psi

B)

P_iV_i = P_fV_f

P_f = (V_i/V_f)P_i = (1.013*10⁵ N/m²)/0.90 = 1.1256*10⁵ N/m²