One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the -gure- The change in internal energy of the gas during the transition is-

The correct option is C –20KJ n = 1, γ [diatomic] = 7/5 Δ U = nC_v Δ T [always] = nC_v [T_2 T_1] = n R/γ 1[T_2 T_1] = nRT_2 nRT_1/γ 1 = P_2 V_2 ...

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Byju's Answer

Standard XII

Chemistry

Entropy for a Reversible Process

One mole of a...

Question

One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the ﬁgure. The change in internal energy of the gas during the transition is:

–12KJ

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20KJ

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–20KJ

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20J

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Solution

The correct option is C
–20KJ

$n = 1, γ [d i a t o m i c] = \frac{7}{5} Δ U = n C_{v} Δ T [a l w a y s] = n C_{v} [T_{2} - T_{1}] = n \frac{R}{γ - 1} [T_{2} - T_{1}] = \frac{n R T_{2} - n R T_{1}}{γ - 1} = \frac{P_{2} V_{2} - P_{1} V_{1}}{γ - 1} = \frac{2 \times 10^{3} \times 6 - 5 \times 10^{3} \times 4}{\frac{7}{5} - 1} = \frac{10^{2}}{2} \times 5 [12 - 20] \Rightarrow - 20 K J C_{V} = \frac{R}{γ - 1} C_{p} = \frac{γ R}{γ - 1}$

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One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the ﬁgure. The change in internal energy of the gas during the transition is:

The correct option is C –20KJ n=1,γ[diatomic]=75ΔU=nCvΔT[always]=nCv[T2−T1]=nRγ−1[T2−T1]=nRT2−nRT1γ−1=P2V2−P1V1γ−1=2×103×6−5×103×475−1=1022×5[12−20]⇒−20KJCV=Rγ−1Cp=γRγ−1