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1st Law of Thermodynamics

In a thermody...

Question

In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by $T Δ X$ , where $T$ is temperature of the system and $Δ X$ is the infinitesimal change in a thermodynamic quantity $X$ of the system. For a mole of monatomic ideal gas $X = \frac{3}{2} R ln (\frac{T}{T_{A}}) + Rln (\frac{V}{V_{A}}) .$ Here, $R$ is gas constant, $V$ is volume of gas, $T_{A}$ and $V_{A}$ are constants.

The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.

List-I List-II

(i)
Work done by the system in process $1 \to 2 \to 3$
(P)
$\frac{1}{3} {R T}_{0} ln 2$

(ii)
Change in intermal energy in process $1 \to 2 \to 3$
(Q)
$\frac{1}{3} {R T}_{0}$

(iii)
Heat absorbed by the system in process $1 \to 2 \to 3$
(R)
${R T}_{0}$

(iv)
Heat absotbed by the system in process $1 \to 2$
(S)
$\frac{4}{3} {R T}_{0}$

(T)
$\frac{1}{3} {R T}_{0} (3 + ln 2)$

(U) $\frac{5}{6} {R T}_{0}$

If the process carricd out on one mole of monoatomic ideal gas is as shown in figure in the PV-diagram with $P_{0} V_{0} = \frac{1}{3} {R T}_{0}$ , the correct match is,

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

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

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$I \to S, I I \to T, I I I \to Q, I V \to U$

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

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Solution

The correct option is B
$I \to P, I I \to R, I I I \to T, I V \to P$

(1) $W_{1 - 2 - 3} = \frac{{R T}_{0}}{3} ln (\frac{V_{2}}{V_{1}}) + 0 = \frac{{R T}_{0}}{3} l n 2$

(2) $Δ U = \frac{3}{2} R (T_{0} - \frac{T_{0}}{3}) = {R T}_{0}$

(3) $Δ Q = \frac{{R T}_{0}}{3} m 2 + {R T}_{0}$
(4) $Δ Q = \frac{{R T}_{0}}{3} ln (2)$

$I \to P, I I \to R,$ III $\to T, I V \to P$

Process $1 \to 2$ is isothermal (temprature constant)
Process $2 \to 3$ is isochoric (volume constant)

$(1)$ Work done in $1 \to 2 \to 3$
$W = W_{1 \to 2} + W_{2 \to 3}$

$= n R T l n \frac{V_{f}}{V_{i}} + W_{2 \to 3}$

$= \frac{R T_{o}}{3} l n \frac{2 V_{o}}{V_{o}} + 0$

$W = \frac{R T_{o}}{3} l n 2$ $\Rightarrow$ option P
(ii) $Δ U$ in $1 \to 2 \to 3$
$Δ U = \frac{f}{2} n R (T_{f} - T_{i})$
$= \frac{3}{2} R (T_{o} - \frac{T_{o}}{3})$

$= \frac{3}{2} R (\frac{2 T_{o}}{3})$ $∴ Δ U = R T_{o} \Rightarrow$ option R

(iii) For any system, first law of thermodynamics for $1 \to 2 \to 3$
$Δ Q = Δ U + Δ W$
$Δ Q = R T_{o} + \frac{R T_{o}}{3} l n 2$
$Δ Q = \frac{R T_{o}}{3} (3 + l n 2) \Rightarrow$ option T

(iv) For process $1 \to 2$ (isothermal)
$Δ Q = Δ U + W$
$= \frac{f}{2} n R (T_{f} - T_{i}) + n R T l n (V_{f} / V_{i})$
$= 0 + R (\frac{T_{o}}{3}) l n (\frac{2 V_{o}}{V_{o}})$
$Δ Q = \frac{R T_{o}}{3} l n 2 \Rightarrow$ option P

Hence the correct option is $I \to P, I I \to R, I I I \to T, I V \to P$

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Similar questions

In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by $T Δ X$ , where $T$ is temperature of the system and $Δ X$ is the infinitesimal change in a thermodynamic quantity $X$ of the system. For a mole of monatomic ideal gas $X = \frac{3}{2} R ln (\frac{T}{T_{A}}) + Rln (\frac{V}{V_{A}}) .$ Here, $R$ is gas constant, $V$ is volume of gas, $T_{A}$ and $V_{A}$ are constants.

The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.

	List-I		List-II
(i)	Work done by the system in process $1 \to 2 \to 3$	(P)	$\frac{1}{3} {R T}_{0} ln 2$
(ii)	Change in intermal energy in process $1 \to 2 \to 3$	(Q)	$\frac{1}{3} {R T}_{0}$
(iii)	Heat absorbed by the system in process $1 \to 2 \to 3$	(R)	${R T}_{0}$
(iv)	Heat absotbed by the system in process $1 \to 2$	(S)	$\frac{4}{3} {R T}_{0}$
		(T)	$\frac{1}{3} {R T}_{0} (3 + ln 2)$
		(U)	$\frac{5}{6} {R T}_{0}$

If the process carricd out on one mole of monoatomic ideal gas is as shown in figure in the PV-diagram with $P_{0} V_{0} = \frac{1}{3} {R T}_{0}$ , the correct match is,

Q. In a thermodynamic process on an ideal monoatomic gas, the infinitesimal heat absorbed by the gas is given by

T Δ X

, where

T

is temperature of the system and

Δ X

is the infinitesimal change in a thermodynamic quantity

X

of the system. For a mole of monoatomic ideal gas

X = \frac{3}{2} R ln (\frac{T}{T_{A}}) + R ln (\frac{V}{V_{A}})

. Here,

V

is volume of gas,

T_{A}

and

V_{A}

are constants. The List - I below gives some quantities involved in a process and List - II gives some possible values of these quantities.

\begin{matrix} List - I & List - II (i) Work done by the system in process 1 \to 2 \to 3 & (P) \frac{1}{3} R T_{0} ln 2 (ii)Change in internal energy in process 1 \to 2 \to 3 & (Q) \frac{1}{3} R T_{0} (iii)Heat absorbed by the system in process 1 \to 2 \to 3 & (R) R T_{0} (iv) Heat absorbed by the system in process 1 \to 2 & (S) \frac{4}{3} R T_{0} (T) \frac{1}{3} R T_{0} (3 + ln 2) (U) \frac{5}{6} R T_{0} \end{matrix}

If the process carried out on one mole of monoatomic ideal gas is as shown in figure in the PV diagram with

P_{0} V_{0} = \frac{1}{3} R T_{0}

. The correct match is-

Q. In a thermodynamic process on an ideal monoatomic gas, the infinitesimal heat absorbed by the gas is given by

T Δ X

, where

T

is temperature of the system and

Δ X

is the infinitesimal change in a thermodynamic quantity

X

of the system. For a mole of monoatomic ideal gas

X = \frac{3}{2} R ln (\frac{T}{T_{A}}) + R ln (\frac{V}{V_{A}})

. Here,

R

is gas constant,

T_{A}

and

V_{A}

are constants. The List - I below gives some quantities involved in a process and List - II gives some possible values of these qunatitites.

\begin{matrix} List - I & List - II (i) Work done by the system in process 1 \to 2 \to 3 & (P) \frac{1}{3} R T_{0} ln 2 (ii)Change in internal energy in process 1 \to 2 \to 3 & (Q) \frac{1}{3} R T_{0} (iii)Heat absorbed by the system in process 1 \to 2 \to 3 & (R) R T_{0} (iv) Heat absorbed by the system in process 1 \to 2 & (S) \frac{4}{3} R T_{0} (T) Δ Q = R T_{0} + \frac{R T_{0}}{3} ln 2 \end{matrix}

If the process carried out on one mole of monoatomic ideal gas is as shown in the

T V

- diagram with

P_{0} V_{0} = \frac{1}{3} R T_{0}

. The correct match is -

The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.

	List-I		List-II
(i)	Work done by the system in process $1 \to 2 \to 3$	(P)	$\frac{1}{3} {R T}_{0} ln 2$
(ii)	Change in intermal energy in process $1 \to 2 \to 3$	(Q)	$\frac{1}{3} {R T}_{0}$
(iii)	Heat absorbed by the system in process $1 \to 2 \to 3$	(R)	${R T}_{0}$
(iv)	Heat absotbed by the system in process $1 \to 2$	(S)	$\frac{4}{3} {R T}_{0}$
		(T)	$\frac{1}{3} {R T}_{0} (3 + ln 2)$
		(U)	$\frac{5}{6} {R T}_{0}$

If the process carricd out on one mole of monoatomic ideal gas is as shown in figure in the PV-diagram with $P_{0} V_{0} = \frac{1}{3} {R T}_{0}$ , the correct match is,

Q. One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the

P V

-diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-1 with the corresponding statements in List-II.

List-I	List-II
P. In process I	1. Work done by the gas is zero
Q. In process II	2 Temperature of the gas remains unchanged
R. In process III	3 No heat is exchanged between the gas and its surroundings
S. In process IV	4 Work done by the gas is $6 P_{0} V_{0}$

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