A tire gauge measures pressure in excess of atmospheric pressure, while P_t and

Question

A tire gauge measures pressure in excess of atmospheric pressure, while P_t and P_f used in the ideal gas law are absolute pressures. So, adding atmospheric pressure to the measured value of P_t at 32 psi = 2.206 times 10^N/m^2, gives the absolute pressure: P_i = (1.013 times 10^5 N/m^2) + (2.206 times 10^5 N/m^2) = 3.219 times 10^5 N/m^2 Substituting in the two temperatures of T_i and T_f = -15 degree C = 258 K leads to: P_f = 258 K/293 K P_i = 1.013 times 10^5 NB/m^2 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 l psi = 6895 N/m^2, the gauge pressure P_ = [P_f - 1.013 times 10^5 N/m^2] 1 psi/6895 N/m^2 = 31.9941 psi What is the pressure in the pump when you push the piston down as described? the initial volume v_i inside the pump is reduced to a final value of V_f = 0.90 v_it thereby changing the pressure. Using the ideal gas law to relate the initial pressure and volume to the final pressure and volume we find: P_i V_i = nRT = P_f V_f set v_f = 0.90 V_i and obtain P_f P_i/0.90 = 1.013 times 10^5 N/m^2/0.90 = 1.1255 times 10^5 N/m^2 Finalize Are the results of the problem consistent with your experience in observing the change in tire pressure when the weather turns colder?

Explanation / Answer

The first one answer is

Pf = (Tf/Ti)*(Pi)

Pf = ((258K)/(293K))*(3.219*10^5)

Pf = 2.834x105 N/m2

The gauge pressure is

Pgauge = (2.834x105 N/m2)*(1 psi / 6895 N/m2) = 41.41 psi.

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