Two mole ideal diatomic gas is heated according to path AB and AC

Question

Two mole ideal diatomic gas is heated according to path AB and AC. If the temperature of state B and state C are equal. Calculate $\frac{q_{A C}}{q_{A B}} \times \frac{7}{6}$ (Assume ideal diatomic gas at low temperature)

Smart Achievers · Accepted Answer

(1)

$A B Isobaric$ $C_{p, m} = \frac{7 R}{2}$

$A C polytropic process$ $P = K V$

$P V^{- 1} = K$

$C_{p, m} = C_{v, m} + \frac{R}{1 - x}$ $[x = - 1] = \frac{5 R}{2} + \frac{R}{2} = 3 R$

$\frac{q_{A C}}{q_{A B}} = \frac{\int_{T_{A}}^{T_{C}} n C_{m} d T}{\int_{T_{A}}^{T_{B}} n C_{m} d T}$ $= \frac{2 \times 3 R \times 2 (T_{C} - T_{A})}{2 \times 7 R (T_{B} - T_{A})} = \frac{6}{7}$ $\Rightarrow \frac{6}{7} \times \frac{7}{6} = 1$

Solution

Q.
Two mole ideal diatomic gas is heated according to path AB and AC. If the temperature of state B and state C are equal. Calculate $\frac{q_{A C}}{q_{A B}} \times \frac{7}{6}$ (Assume ideal diatomic gas at low temperature)

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Solution

Q. Two mole ideal diatomic gas is heated according to path AB and AC. If the temperature of state B and state C are equal. Calculate qACqAB×76 (Assume ideal diatomic gas at low temperature)

Ans. (1) AB Isobaric Cp,m=7R2 AC polytropic process P=KV PV-1=K Cp,m=Cv,m+R1-x [x=-1]=5R2+R2=3R qACqAB=∫TATCnCm dT∫TATBnCm dT=2×3R×2(TC-TA)2×7R(TB-TA)=67⇒67×76=1

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Q.
Two mole ideal diatomic gas is heated according to path AB and AC. If the temperature of state B and state C are equal. Calculate $\frac{q_{A C}}{q_{A B}} \times \frac{7}{6}$ (Assume ideal diatomic gas at low temperature)