A sample of one mole of an ideal gas at 1.0 atm and 300 K. $(C_{p} = 7 / 2 R)$ is put through the following:

a) Constant volume heating to twice its initial temperature
b) Reversible adiabatic expansion back to its initial temperature
c) Reversible isothermal compression to 1.0 atm.
(take $l n 2 = 0.3, R = \frac{25}{3}$ in SI units)

Match the following:
$\begin{matrix} Quantity & Value P & w_{b} & I & 0 k J Q & w_{c} & I I & - 6.25 k J R & q_{t o t} & I I I & 4.375 k J S & Δ E_{t o t} & I V & 1.875 k J \end{matrix}$

R-I, S-II

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R-II, S-III

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R-III, S-IV

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R-IV, S-I

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Solution

The correct option is D R-IV, S-I
$\begin{matrix} S t a t e & P & V & T 1 & 1 a t m & V_{1} & 300 K 2 & p_{2} & V_{2} = V_{1} & 600 K 3 & p_{3} & V_{3} & 300 K 4 & 1 a t m & V_{4} = V_{1} & 300 K \end{matrix}$

$w_{a} = 0 Δ E_{a} = n C_{v} (T_{2} - T_{1}) \Rightarrow Δ E_{a} = 1 \times \frac{5}{2} \times \frac{25}{3} \times (600 - 300) = 6250 J ∴ q_{a} = Δ E_{a} = 62.5 J \frac{T_{3}}{T_{2}} = (\frac{V_{2}}{V_{3}})^{(γ - 1)} γ - 1 = \frac{R}{C_{V}} = \frac{R}{2.5 R} = 0.4 ∴ \frac{300}{600} = (\frac{25}{V_{3}})^{0.4} \Rightarrow V_{3} = 2^{2.5} \times 25 = 100 \sqrt 2 L Δ E_{b} = n C_{V} Δ T = 1 \times 5 / 2 R (300 - 600) = - 750 R = - 6250 J w_{b} = Δ E_{b} = - 6.25 k J w_{c} = - n R T l n \frac{V_{4}}{V_{3}} \Rightarrow w_{c} = - 1 \times 25 / 3 \times 300 \times l n \frac{25}{100 \sqrt 2} \Rightarrow w_{c} = 4375 J Δ E_{c} = 0 q_{c} = - w_{c} = - 4375 J q_{t o t} = q_{a} + q_{b} + q_{c} = 6250 + 0 - 4375 = 1875 J w_{t o t} = w_{a} + w_{b} + w_{c} = 0 + (- 6250) + (4375) = - 1875 J Δ E_{t o t} = Δ E_{a} + Δ E_{b} + Δ E_{c} = 6250 + (- 6250) + 0 = 0$

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Similar questions

List - I (Process)	List - II
(I) Reversible isothermal expansion of an ideal gas.	(P) $Δ S_{s y s} = 0$
(II) Reversible adiabatic compression of a real gas.	(Q) $Δ H = Δ U = Δ S_{total} = 0$
(III) Adiabatic free expansion	(R) $Δ S_{total} > 0$
(IV) Isothermal free expansion	(S) $q = 0$
	(T) $Δ S_{surr} = 0$
	(U) $Δ S_{s y s} > 0$

Which of the following options has the correct combination considering List - I and List - II.

List - I (Process)	List - II
(I) Reversible isothermal expansion of an ideal gas.	(P) $Δ S_{s y s} = 0$
(II) Reversible adiabatic compression of a real gas.	(Q) $Δ H = Δ U = Δ S_{total} = 0$
(III) Adiabatic free expansion.	(R) $Δ S_{total} > 0$
(IV) Isothermal free expansion	(S) $q = 0$
	(T) $Δ S_{surr.} = 0$
	(U) $Δ S_{s y s} > 0$

Which of the following options has the incorrect combination considering List - I and List - II.

Q. Calculate q, for the reversible isothermal expansion of one mole of an ideal gas at

127^{\circ} C

from a volume of

10 {dm}^{3}

20 {dm}^{3}

Take

ln 2 = 0.3

and

R = 8.3 {J K}^{- 1} {mol}^{- 1}

Column-1 represents the process
Column-2 represents the $w = Δ U$ for that process and
Column-3 represents the work done expression for that process.
$\begin{matrix} C o l u m n I & C o l u m n 2 & C o l u m n 3 (I) & A d i a b a t i c e x p a n s i o n & (i) & 0 & (P) & w = q (I I) & I r r e v e r s i b l e i s o t h e r m a l c o m p r e s s i o n & (i i) & P o s i t i v e & (Q) & w = - q (I I I) & R e v e r s i b l e i s o t h e r m a l c o m p r e s s i o n & (i i i) & N e g a t i v e & (R) & w = 0 (I V) & F r e e e x p a n s i o n & (i v) & q & (S) & w = ? U \end{matrix}$
Which set is correct for the bursting of tyre?

Q. Calculate q, for the reversible isothermal expansion of one mole of an ideal gas at

127^{\circ} C

from a volume of 10 dm

^{3}

to 20 dm

^{3}

Take ln2 = 0.3 and R = 8.3

J K^{- 1} m o l^{- 1}

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