For 1 mol of a gas, the Ideal Gas Law can be stated in the following form, which
ID: 525934 • Letter: F
Question
For 1 mol of a gas, the Ideal Gas Law can be stated in the following form, which expresses pressure, P, as a function of volume, V, and temperature. T. P(V, T) = 8.314 T/V In this formula, the units of pressure are Pascals, the units of temperature are degrees Kelvin, and the units of volume are cubic meters. a) Calculate P_v(20, 300) and explain what this number means in terms of the relationship between pressure, volume, and temperature. b) Suppose that volume and temperature are both changing in such a way that when the volume increases by 1 cubic meter, the temperature increases by 2 degrees Kelvin. Under these circumstances, find the instantaneous rate of change of pressure when the volume is 20 cubic meters and the temperature is 300 degrees Kelvin. c) The graph of P(V, T) = 8314 T/V is a surface in three-dimensional space. Find a set of parametric equations for the tangent line to this surface in the "V direction" at the point on the surface corresponding to (20, 300).Explanation / Answer
IDEAL GAS EQUATION FOR IMOLE OF A GAS IS
PV = RT
P = RT/V
= 8.314T/V
WHERE V IS MEASURED IN CUBIC METERS
T IS MEASURED IN KELVIN DEGREES
AND P IS MEASURED IN PASCALS.
a) GIVEN THAT V = 20 AND T=300, THEN
Pv = 8.314 * 300 / 20 = 124.71 PASCALS
THIS MEANS THAT 124.71 PASCALS OF PRESUURE IS REQUIRED TO RAISE THE TEMPERATURE
OF A FIXED VOLUME OF 20 cm3 OF 1 MOLE OF GAS FROM 2730 K TO 3000 K.
b) IN THIS CASE, WE HAVE TO TAKE
P1 = 124.71
WHEN THE VOLUME IS INCREASED BY 1 CUBIC METER AND THE TEMPERATURE IS RAISED BY 2 DEGREES KELVIN,
THEN , V= 21 AND T = 302
P2 = 8.314 * 302 / 21 = 119.56
THEN THE INSTANTANEOUS CHANGE IN PRESSURE OF 124.71 - 119.56 = 5.15
THEN THE INSTANTANEOUS RATE OF CHANGE OF PRESSURE IS 5.15 / 124.71 = 0.041
c) SET OF PARAMETERS ( P=124.71 , V=20, T=300)
AND ( P=119.56 , V=21, T=302)
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