A Constant-Volume Gas Thermometer. ?An experimenter using a gas thermometer found the pressure at the triple point of water (0.01°C) to be and the pressure at the normal boiling point were what pressure would the experimenter have measured at (As we will learn in Section 18.1, Eq. (17.4) is accurate only for gases at very low density.)

Solution 9E A Constant-Volume Gas Thermometer. T2/T 1 P /P2 1 Here in above equation the ratio of two temperature in kelvins is equal to ratio of corresponding pressures in constant volume gas thermometer. Consider a graph of P against T. This, we are told is a straight line. We can find the value of T at which P = 0 by finding the slope between the two given points and then extrapolating downwards. The pressure to temperature linear relation is P 2 P =1(T T2) 1 Here P and2P are t1e initial and final pressure,T and T are corre1ponding2 temperatures and is pressure coefficient of temperature. Substitute values P = 4.80 × 10 Paand P = 6.50 × 10 Pa. 4 1 2 0 0 0.01 C for T an1 100 C forT ,to fin2 value of . (6.50 × 10 Pa) (4.80 × 10 Pa) = ( 100 C 0 C ) 0 4 4 0 0 = (6.50 × 10 Pa) (4.80 × 10 Pa)/( 100 C 0 C ) = 170 Pa/ C0 0 Hence the value of is = 170 Pa/ C . Using the upper data point at 100 C to avoid troublesome fractions, we can write 4 0 0 change in temp for pressure to be zero is (0 6.5 × 10 Pa)/170 Pa/ C = 382.3 C so absolute zero is at (100 C 382.3 C = 282.3 C. 0