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$\bullet$$\bullet$ Figure 15.43 shows a $p V$ diagram for 0.0040 mole of ideal $\mathrm{H}_{2}$ gas. The temperature of the gas does not change during segment bc. (a) What volume does this gas occupy at point $c$ ? (b) Find the temperature of the gas at points $a, b,$ and $c .$ (c) How much heat went into or out of the gas during segments $a b, c a,$ and $b c ?$ Indicate whether theheat has gone into or out of the gas. (d) Find the change in the internal energy of this hydrogen during segments $a b, b c,$ and $c a .$ Indicate whether the internal energy increased or decreased during each of these segments.

(a) $V_{c}=0.80 L$(b) $T_{c}=1220 \mathrm{k}$(c) $Q=76 J$(d)$\Delta u=76 \mathrm{~J}$

Physics 101 Mechanics

Chapter 15

Thermal Properties of Matter

Temperature and Heat

The First Law of Thermodynamics

The Second Law of Thermodynamics

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(a) One-third of a mole of…

The figure shows speed diagram for 0 point double 040 mole of ideal hydrogen gas, and it is given in the problem that, in the segment b c, temperature is a constant now in part, a of this problem. We want to calculate the volume of the gas 8 point c now in the segment bc. Temperature is constant, so we can write the ideal gas equation in this form, their volume at c times. Pressure at c is equal to volume at b times pressure at b. So pressure volume at c is equal to volume at b times pressure at b, upon pressure at c now, by putting different values here, volume at b is 0.20 liter multiply. Pressure at b is 2 atmosphere upon pressure. R c is 0.50 atmosphere, so we have. The value of v c is equal to 0.80 litter. So this is the volume of gas 8 point c. Now, in the second part of this problem, we are going to calculate the temperature at points a b and c. Now, first take point a 8 point: a temperature 8 point: a is equal to pressure at point a times. Volume 8 point a upon an times r, now by putting different values here here, pressure at point: a is 0.5 atmosphere that is equal to 0.5 times 1013. Double 0 atmosphere and volume 8 point: a is 0.2 liter that is equal to 0.2 multiply 10 to the power minus 3 meter cube upon number of moles. Here is 0.004 mole, and the value of r is 8.314 joule per mole per kelvin. So we get temperature 8 point a is equal to 305 kelvin. Now we want to calculate temperature at point b in the segment. A b volume of the gas is a constant, so we have temperature upon pressure is equal to constant in the segment by using real gas equation. We have p v is equal to n r, or we can write it as temperature. Upon pressure is equal to volume. Upon or here in the segment volume n and are our constants, so we have tmperature upon p, is equal to a constant number so for the segment we can write this as temperature at point b, upon pressure at point b is equal to temperature 8 point a Unpressure at point a or we can express this as temperature at b is equal to pressure at b, upon pressure at a multiply temperature at now, by putting different values here, pressure at b is 2 atmosphere upon pressure at a is 0.5 atmosphere and the temperature. At point is 305 kelvin, so we get temperature at b equal to 1220 kelvin. Now we want to calculate temperature at point c, as it is given in the problem that sigment b c is an isotherm, so here temperature is constant. This shows that temperature at c is equal to temperature at b, so temperature at c will be equal to 1220 kelvin. Now, in the next part of this problem, we want to calculate the heat released are absorbed in each segment of the diagram. Now let us take the first segment here. A b here the heat released or absorbed is equal to n times, molars specificated, constant volume times delta, t hydrogenatomic gas, and we have considered hydrogen as a real gas than the value of molar specific heat at constant value. Will be equal to 5 upon 24, so we have q equal to n times 5 upon 2 r times delta t now why putting different values we have q equal to here, n is 0.004 mole multiply 5.2 and r, that is 8.314 joule per kelvin, and the Change in temperature is temperature at b, that is 1220 kelvin minus temperature 8, a that is 305 kelvin, so we have c equal to 76 joule. Here we have final answer in positive. This shows that heat is added to the system, so here heat flows into the gas. Now we want to calculate heat in the segment c again. Taking the relation q is equal to n times c p times delta t here in the segment car pressure is constant. As for diatomic ardial guess, we have c p equal to 7 upon 2 r, so we have q equal to n into 7 upon 2 r into delta t now by plugging different values. Here n is 0.004 mole into 7 upon 2. Here r is 8.314 joule per mole per kelvin, and a change in temperature is your final temperature is temperature at, and that is 305 kelvin minus temperature 8 c, that is 1220 kelvin. So we get heat in this segment equal to minus 107. Joule here negative sign shows that heat is flowing out of the gas. Now we want to calculate the heat in the segment b c in the segment bc. Temperature is constant. So by using fossila of thermo dynamics, we have c equal to w plus delta. U here, temperature is constant, so delta t is equal to 0. This shows that delta, u s equal to 0, so we have c equal to w s. Work done in agethermic process is equal to n times r t into natural logarithm of final volume. Her final volume is vere upon initial volume that is v 8 b now, by putting the value of w. Here we have q, equal to n r t, l n of v c upon v b now by substituting different values. Here, n is 0.004 mole. R is 8.314 joule per mole per kelvin, and the temperature here is 1220 kelvin into natural log of volume. At c, that is 0 point a liter upon volume at b, that is 0.2 liter, so we get q in this segment is equal to 56 joule. Here the answer is positive. This shows that heat flows into the gas. Now in the last part of this problem, we want to calculate change in internal energy in each segment. Now, first take the segment a b as we have the relation for change in internal energy. That delta? U is equal to n times c v times delta t here c v is 5.2 r into delta t pot. Now, by plugging different values, we have m equal to 0.004. More multiply 5.2 here. R is and .314 joule per mole per kelvin and the change in temperature in the segment is temperature at b, which is 1220 kelvin minus temperature at a that is 305 kelvin. So we get change internal energy equal to 76 joulenow in the segment b c. In the segment bc, temperature is a constant, so we have delta t equal to 0, as the relation for change in internal energy is equal to n times c v times delta t here, delta t is equal to 0. That'S why change in internal energy is equal to 0 poinow in the segment c, we have change in internal energy equal to 10 times molar specific heat at constant volume times change in temperature, which is equal to n, and here c v is 5.2 r into changing Temperature now by plugging different values, here n is 0.004 mole into 5.2. Here r is 8.314 joule per mole per kelvin and changing temperature, that is, temperature, which is 305 kelvin minus temperature 8 c, which is 1220 kelvin. So we get change in internal energy. In this statement is equal to minus 76 joule there's the end of the problem watching.

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