Question 130: Thermodynamics & KTG

Question

List I describes thermodynamic processes in four different systems. List II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system due to the process.

List-I
List-II

(I) $10^{-3} \mathrm{~kg}$ of water at $100^{\circ} \mathrm{C}$ is converted to steam at the same temperature, at a pressure of $10^{5} \mathrm{~Pa}$. The volume of the system changes from $10^{-6} \mathrm{~m}^{3}$ to $10^{-3} \mathrm{~m}^{3}$ in the process. Latent heat of water $=2250\, \mathrm{~kJ} / \mathrm{kg}$.
(P) $2 \mathrm{~kJ}$

(II) $0.2$ moles of a rigid diatomic ideal gas with volume $V$ at temperature $500 \mathrm{~K}$ undergoes an isobaric expansion to volume $3 \mathrm{~V}$. Assume $R=8.0 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$.
(Q) $7 k J$

(III) One mole of a monatomic ideal gas is compressed adiabatically from volume $V=\frac{1}{3} \mathrm{~m}^{3}$ and pressure $2 \mathrm{kPa}$ to volume $\frac{V}{8}$.
(R) $4 \mathrm{~kJ}$

(IV) Three moles of a diatomic ideal gas whose molecules can vibrate, is given $9 \mathrm{~kJ}$ of heat and undergoes isobaric expansion.
(S) $5 \mathrm{~kJ}$

(T) $3 \mathrm{~kJ}$

Which one of the following options is correct?

Accepted Answer

$$U = ML - P\Delta V$$
$$ = {10^{ - 3}} 	imes 2250 - {10^2}kP 	imes ({10^{ - 3}} - {10^{ - 6}})$$ m
3
= 2.25 kJ $$-$$ 0.1 kJ
= 2.15 kJ
I $$	o$$ P
(II) $${C_V} = {{5R} \over 2}$$ (rigid diatomic)
For isobaric expansion
$$V \propto T$$
$${{{V_1}} \over {{V_2}}} = {{{T_1}} \over {{T_2}}} \Rightarrow {V \over {3V}} = {{500} \over {{T_2}}} \Rightarrow {T_2} = 1500\,K$$
$$\Delta U = n{C_V}\Delta T = 0.2 	imes {{5 	imes 8} \over 2} 	imes (1500 - 500)$$ J = 4 kJ
II $$	o$$ R
(III) Adiabatic expansion $$\left( {\gamma  = {5 \over 3}} ight)$$
$${P_1}V_1^\gamma  = {P_2}V_2^\gamma  \Rightarrow $$ (2 kPa) $$ 	imes V_0^{5/3} = {P_2} 	imes {\left( {{{{V_0}} \over 8}} ight)^{5/3}}$$
$${P_2} = 64$$ kPa
$$\Delta U = n{C_V}\Delta T$$
$$ = {{3nR\Delta T} \over 2} = {3 \over 2}({P_2}{V_2} - {P_1}{V_1}) = {3 \over 2} 	imes \left( {64 	imes {1 \over {3 	imes 8}} - 2 	imes {1 \over 3}} ight)$$
$$ = {3 \over 2} 	imes \left( {{8 \over 3} - {2 \over 3}} ight) = 3$$ kJ
III $$	o$$ T
(IV) For isobaric expaision,
$$\Delta U = n{C_V}\Delta T = {7 \over 2}\,nR\Delta T$$
$$\Delta Q = n{C_P}\Delta T = {9 \over 2}\,nR\Delta T$$
$${{\Delta U} \over {\Delta Q}} = {7 \over 9}$$
$$\Delta U = {7 \over 9}\Delta Q = 7$$ kJ
IV $$	o$$ Q

Answer

I $\rightarrow$ T, II $\rightarrow$ R, III $\rightarrow$ S, IV $\rightarrow$ Q

Answer

I $\rightarrow$ S, II $\rightarrow$ P, III $\rightarrow$ T, IV $\rightarrow$ P

Answer

I $\rightarrow$ P, II $\rightarrow$ R, III $\rightarrow$ T, IV $\rightarrow$ Q

Answer

I $\rightarrow$ Q, II $\rightarrow$ R, III $\rightarrow$ S, IV $\rightarrow$ T

Thermodynamics & KTG

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