This is from my Fluids class, Im assuming it should be under Advanced physics? 1. The problem statement, all variables and given/known data A- The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25C the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m^3, determine the pressure rise in the tire when the air temperature in the tire rises to 50C. Also determine the air that must be bled off to restore the original pressure at this temperature. Assume the atmospheric pressure is 100 kPa 2. Relevant equations P= (rho) RT (rho)=m/V (P_1 V_1)/T_1 = (P_2 V_2)/T_2 according to my book, for the compression and expansion coefficients (alpha and beta) (delta)(rho) = -(beta)*(rho)*(delta)T (delta)(rho) = (alpha)*(rho)*(delta)P 3. The attempt at a solution so I can easily find the pressure at 50 degrees (I convert to Kelvin first) so at 50C the pressure is 231.51 kPa so then (delta)P=21.51 kPa then using the forumlas for the coefficients of expansion and compression: assuming rho stays constant (since there is no more mass added to the tire and the tire isnt changing volume (therefore mass and volume are constant therefore rho should be constant?) in those formulas Beta would be 1/(delta)T and alpha would be 1/(delta)P which are .04 and .0464 respectively and then to be honest I am completely lost so I just scribbled stuff and for the third part (how much air should be left out) I used the formula dleta rho=-beta*rho*(delta)T I replaced delta rho with delta m/V (because the mass will change but the volume will stay constant) and then I solved for delta m which gave me a result of .025(rho) but rho = m_1/v and delta m = m_1 - m_0 so after playing around I got that m-1 = 1/V and I finally got 41 kg, but I know this is totally wrong I doubt theres 45 kg worth of air inside the tire. . . ====================== 1. The problem statement, all variables and given/known data B- The density of seawater at a free surface where pressure is 92kPa is approxdimately 1030 kg/m^3. Taking the bulk modulus of elasticity of seawater (kappa) to be 2.34x10^9 N/m^2 and expressing variation of pressure with dept z as dP= (rho)*g*dz determine the density and pressure at a depth of 2500 m. Disregard temperature 2. Relevant equations 3. The attempt at a solution getting the specific gravity for sea water SG= 1030/1000 = 1.030 (rho)_0 = 1030 kg/m^3 @ 98 kPa (kappa)= 2.34x10^9 (kappa)= dP/(d(rho)/(rho)) <- solve for dP and equate with dP= (rho)*g*dz and then Im lost. . .