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Question: $\text{one mole monoatomic}$ $\text{gas, heat Capacity at}$ $\text{Point A=?}$ $\text{A = equidis...

one mole monoatomic\text{one mole monoatomic}

gas, heat Capacity at\text{gas, heat Capacity at}

Point A=?\text{Point A=?}

A = equidistant from\text{A = equidistant from}

Both Intersections\text{Both Intersections}

Answer

R/2

Explanation

Solution

The process is represented by a straight line in the V-T diagram. Let the equation of the line be V=mT+cV = mT + c. The line intersects the T-axis at point C (where V=0) and the V-axis at point B (where T=0). From the figure, the line has a negative slope and positive intercepts on both axes. Let the T-intercept be T0T_0 and the V-intercept be V0V_0. The equation of the line passing through (T0,0)(T_0, 0) and (0,V0)(0, V_0) is TT0+VV0=1\frac{T}{T_0} + \frac{V}{V_0} = 1. So, V=−V0T0T+V0V = -\frac{V_0}{T_0} T + V_0. Here m=−V0T0m = -\frac{V_0}{T_0} and c=V0c = V_0. The intersections are C=(T0,0)C = (T_0, 0) and B=(0,V0)B = (0, V_0). Point A = (TA,VA)(T_A, V_A) lies on the line, so VA=−V0T0TA+V0V_A = -\frac{V_0}{T_0} T_A + V_0. Point A is equidistant from B and C. The distance squared from A to B is AB2=(TA−0)2+(VA−V0)2=TA2+(VA−V0)2AB^2 = (T_A - 0)^2 + (V_A - V_0)^2 = T_A^2 + (V_A - V_0)^2. The distance squared from A to C is AC2=(TA−T0)2+(VA−0)2=(TA−T0)2+VA2AC^2 = (T_A - T_0)^2 + (V_A - 0)^2 = (T_A - T_0)^2 + V_A^2. Since AB2=AC2AB^2 = AC^2, we have TA2+(VA−V0)2=(TA−T0)2+VA2T_A^2 + (V_A - V_0)^2 = (T_A - T_0)^2 + V_A^2. From the line equation, VA−V0=−V0T0TAV_A - V_0 = -\frac{V_0}{T_0} T_A. So, TA2+(−V0T0TA)2=(TA−T0)2+VA2T_A^2 + (-\frac{V_0}{T_0} T_A)^2 = (T_A - T_0)^2 + V_A^2. TA2+V02T02TA2=TA2−2TAT0+T02+VA2T_A^2 + \frac{V_0^2}{T_0^2} T_A^2 = T_A^2 - 2T_A T_0 + T_0^2 + V_A^2. V02T02TA2=−2TAT0+T02+VA2\frac{V_0^2}{T_0^2} T_A^2 = - 2T_A T_0 + T_0^2 + V_A^2. Substitute VA=V0(1−TAT0)V_A = V_0 (1 - \frac{T_A}{T_0}). V02T02TA2=−2TAT0+T02+V02(1−TAT0)2\frac{V_0^2}{T_0^2} T_A^2 = - 2T_A T_0 + T_0^2 + V_0^2 (1 - \frac{T_A}{T_0})^2. V02T02TA2=−2TAT0+T02+V02(1−2TAT0+TA2T02)\frac{V_0^2}{T_0^2} T_A^2 = - 2T_A T_0 + T_0^2 + V_0^2 (1 - 2\frac{T_A}{T_0} + \frac{T_A^2}{T_0^2}). V02T02TA2=−2TAT0+T02+V02−2V02TAT0+V02TA2T02\frac{V_0^2}{T_0^2} T_A^2 = - 2T_A T_0 + T_0^2 + V_0^2 - 2 \frac{V_0^2 T_A}{T_0} + \frac{V_0^2 T_A^2}{T_0^2}. 0=−2TAT0+T02+V02−2V02TAT00 = - 2T_A T_0 + T_0^2 + V_0^2 - 2 \frac{V_0^2 T_A}{T_0}. 2TAT0+2V02TAT0=T02+V022T_A T_0 + 2 \frac{V_0^2 T_A}{T_0} = T_0^2 + V_0^2. 2TA(T0+V02T0)=T02+V022T_A (T_0 + \frac{V_0^2}{T_0}) = T_0^2 + V_0^2. 2TAT02+V02T0=T02+V022T_A \frac{T_0^2 + V_0^2}{T_0} = T_0^2 + V_0^2. Assuming T02+V02≠0T_0^2 + V_0^2 \neq 0 (which is true since T0>0T_0 > 0 and V0>0V_0 > 0), we can divide by T02+V02T_0^2 + V_0^2. 2TA/T0=1  ⟹  TA=T0/22T_A/T_0 = 1 \implies T_A = T_0/2. Now find VAV_A: VA=V0(1−TAT0)=V0(1−T0/2T0)=V0(1−1/2)=V0/2V_A = V_0 (1 - \frac{T_A}{T_0}) = V_0 (1 - \frac{T_0/2}{T_0}) = V_0 (1 - 1/2) = V_0/2. So, point A is (T0/2,V0/2)(T_0/2, V_0/2).

The molar heat capacity for a process is given by C=CV+pndVdTC = C_V + \frac{p}{n} \frac{dV}{dT}. For 1 mole of monoatomic ideal gas, n=1n=1 and CV=32RC_V = \frac{3}{2}R. Also pV=RTpV = RT, so p=RTVp = \frac{RT}{V}. C=32R+RTVdVdTC = \frac{3}{2}R + \frac{RT}{V} \frac{dV}{dT}. The process is a straight line in the V-T diagram, so dVdT=m=−V0T0\frac{dV}{dT} = m = -\frac{V_0}{T_0}. At point A, T=TA=T0/2T=T_A = T_0/2 and V=VA=V0/2V=V_A = V_0/2. CA=32R+R(T0/2)(V0/2)(−V0T0)C_A = \frac{3}{2}R + \frac{R (T_0/2)}{(V_0/2)} (-\frac{V_0}{T_0}). CA=32R+RT0V0(−V0T0)C_A = \frac{3}{2}R + \frac{R T_0}{V_0} (-\frac{V_0}{T_0}). CA=32R−RC_A = \frac{3}{2}R - R. CA=12RC_A = \frac{1}{2}R.