List I describes thermodynamic processes in four different systems....

(I)

According to the first law of thermodynamics,

$∆ Q = ∆ U + W ⇒ ∆ U = ∆ Q - W$

$⇒ U = ML - P\DeltaV$

$= {10}^{- 3} \times 2250 - {10}^{2}kP \times \left({10}^{- 3} - {10}^{- 6}\right){m}^{3}$

$= 2 . 25kJ - 0 . 1kJ$

$= 2 . 15kJ$

Therefore, I-P

(II)

For isobaric process,

$\dfrac{{V}_{1}}{{V}_{2}} = \dfrac{{T}_{1}}{{T}_{2}} ⇒ {T}_{2} = 3 \times 500 = 1500K$

Now the change in the internal energy will be,

$∆ U = n{C}_{V} ∆ T = 0 . 2 \times \dfrac{5}{2} \times 8 \times 1000 = 4kJ$

Therefore, II-R

(III)

For adiabatic expansion of monoatomic gas $\left(\gamma = \dfrac{5}{3}\right)$

${P}_{1}{V}_{1}^{\gamma} = {P}_{2}{V}_{2}^{\gamma} ⇒ \left(2kPa\right) \times {V}_{0}^{\dfrac{5}{3}} = {P}_{2} \times \left(\dfrac{{V}_{0}}{8}\right)\dfrac{5}{3}$

$⇒ {P}_{2} = 64kPa$

Now the change in the internal energy will be,

$\DeltaU = n{C}_{v}\DeltaT$

$= \dfrac{3nR\DeltaT}{2} = \dfrac{3}{2}\left({P}_{2}{V}_{2} - {P}_{1}{V}_{1}\right) = \dfrac{3}{2} \times \left(64 \times \dfrac{1}{3 \times 8} - 2 \times \dfrac{1}{3}\right)$

$= \dfrac{3}{2} \times \left(\dfrac{8}{3} - \dfrac{2}{3}\right) = 3kJ$

Therefore, III-T

(IV)

For diatomic gas whose molecules vibrate,

${C}_{V} = \dfrac{6}{2}R$ and ${C}_{P} = \dfrac{8}{2}R$

$⇒ \DeltaU = n{C}_{V}\DeltaT = 3nR\DeltaT$

$\DeltaQ = n{C}_{p}\DeltaT = 4nR\DeltaT$

$\dfrac{\DeltaU}{\DeltaQ} = \dfrac{3}{4}$

$\DeltaU = \dfrac{3}{4} \times 9 = 6 . 75 ≃ 7kJ$

Therefore, IV-Q