You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Work Done

List I descri...

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

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

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

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

(T) $3 kJ$

Which one of the following options is correct?

I \to S, I I \to P, I I I \to T, I V \to P

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I \to Q, I I \to R, I I I \to S, I V \to T

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I \to T, I I \to R, I I I \to S, I V \to Q

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I \to P, I I \to R, I I I \to T, I V \to Q

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $I \to P, I I \to R, I I I \to T, I V \to Q$
$(1)$ As we know that,
$W = P Δ V = 10^{5} (10^{- 3} - 10^{- 6}) J = 100 (10^{- 3} - 10^{6}) kJ = 0.1 kJ$
Also, $Q = m L = 10^{- 3} (2250) = 2.25 kJ$
Thus, we have
$Δ U = Q - W \Rightarrow Δ U = 2.25 - 0.1 = 2.15 kJ ≃ 2 kJ$

$(2)$
For isobaric expansion
$V \propto T$
$\Rightarrow \frac{V_{1}}{V_{2}} = \frac{T_{1}}{T_{2}} \Rightarrow \frac{V}{3 V} = \frac{500}{T_{2}} \Rightarrow T_{2} = 1500 K$
and $Δ U = n C_{v} Δ T$
$= 0.2 \times \frac{5}{2} R \times (1500 - 500) = 0.2 \times 40 (500) = 40 \times 100 = 4 kJ$

$(3)$
For adiabatic expansion, $(γ = \frac{5}{3})$
and,
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\Rightarrow P_{1} (V)^{5 / 3} = P_{2} {(\frac{V}{8})}^{5 / 3}$
$\Rightarrow P_{2} = (8^{5 / 3}) P_{1} = 32 P_{1} = 64 kPa$
Now, we have
$Δ U = \frac{n R (Δ T)}{γ - 1}$
$= \frac{P_{2} V_{2} - P_{1} V_{1}}{γ - 1} = \frac{\frac{64 V}{8} - 2 V}{\frac{5}{3} - 1} = \frac{6 V}{(\frac{2}{3})} = \frac{18 V}{2} = 9 V = 9 \times \frac{1}{3} = 3 kJ$

$(4)$
For isobaric expansion,
$Δ U = n C_{v} Δ T$
$= \frac{5}{2} n R Δ T$
$C_{V} = \frac{f}{2} R = \frac{7}{2} R$
$Q = n C_{P} Δ T = \frac{9}{2} n R Δ T$
$Δ U = n C_{V} Δ T = \frac{7}{2} n R Δ T$
$\frac{Q}{Δ U} = \frac{9}{7}$
$Δ U = \frac{7}{9} Q = \frac{7}{9} \times 9 kJ = 7 kJ$
Hence option $(C)$ is correct.

Suggest Corrections

Similar questions

$n$ moles of a perfect gas undergoes a cyclic process $A B C A$ (see figure) consisting of the following processes:

$A \to B$ : Isothermal expansion at a temperature $T$ so that the volume is doubled $V_{1} t o V_{2}$ and pressure changes $P_{1} t o P_{2}$ .

$B \to C$ : Isobaric compression at pressure $P_{2}$ to initial volume $V_{1}$ .

$C \to A$ : Isochoric change leading to change of pressure from $P_{1} t o P_{2}$ .

Total work done in the complete cycle $A B C A$ is –

Q. Two moles of an ideal monoatomic gas occupies a volume

V

27^{\circ} C

. The gas expands adiabatically to a volume

2 V .

Calculate (a) the final temperature of the gas and (b) change in its internal energy.

Q. An ideal gas initially at

30 K

undergoes an isobaric expansion at

2.50 k P a

. If the volume increases from

1.00 m^{3}

3.00 m^{3}

and

12.5 k J

is transferred to the gas by heat, What is the change in its internal energy?

In a flexible balloon, 2 moles of $S O_{2}$ having an initial volume of $1 k L$ at a temperature of $27^{o} C$ is filled. ( $S O_{2}$ is a linear triatomic gas). The gas is first expanded to thrice its initial volume isobarically and then expanded adiabatically so as to attain its initial temperature. Assuming gas is ideal and $R = \frac{25}{3} J m o l^{- 1} K^{- 1}$ change in internal energy of the gas in the isobaric process is:

Q. Draw the pressure-temperature and volume-temperature diagrams for an isobaric process of expansion, corresponding to

n

moles of an ideal gas at a pressure

P_{0}

, from

V_{0}

2 V_{0}

Join BYJU'S Learning Program

	List -I		List -II
(I)	$10^{- 3} kg$ of water at $100^{\circ} C$ is converted to steam at the same temperature, at a pressure of $10^{5} Pa$ . The volume of the system changes from $10^{- 6}$ to $10^{- 3}$ in the process. Latent heat of water $= 2250 kJ/kg$ .	(P)	$2 kJ$
(II)	$0.2$ moles of a rigid diatomic ideal gas with volume $V$ at temperature $500 K$ undergoes an isobaric expansion to volume $3 V$ . Assume $R = 8.0 {J mol}^{- 1} K^{- 1}$ .	(Q)	$7 kJ$
(III)	One mole of a monoatomic ideal gas is compressed adiabatically from volume $V = \frac{1}{3} m^{3}$ and pressure $2 kPa$ to volume $\frac{V}{8}$	(R)	$4 kJ$
(IV)	Three moles of a diatomic ideal gas whose molecules can vibrate, is given $9 kJ$ of heat and undergoes isobaric expansion.	(S)	$5 kJ$
		(T)	$3 kJ$