Question 1: Thermodynamics & KTG

Question

An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature T
1
, pressure P
1
and volume V
1
and the spring is in its relaxed state. The gas is then heated very slowly to temperature T
2
, pressure P
2
and volume V
2
. During this process the piston moves out by a distance x.
Ignoring the friction between the piston and the cylinder, the correct statements is/are
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l0wnwxbe/ba1320d0-62d3-47c5-aa9e-3b8bd4e64920/38567990-a6dc-11ec-927d-c55e47b0758b/file-1l0wnwxbf.png?format=png)

Accepted Answer

Initially both the compartments has same pressure as they are in equilibrium.
Suppose spring is compressed by x on heating the gas.
Let A be the area of cross-section of piston. As gas is ideal monoatomic, so
${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}{V_2}} \over {{T_2}}}$ ...... (i)
Force on spring by gas = kx
$	herefore$ ${P_2} = {P_1} + {{kx} \over A}$ ...... (ii)
Case I : When ${V_2} = 2{V_1}$, ${T_2} = 3{T_1}$
From Eqn. (i)
${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}(2{V_1})} \over {3{T_1}}} \Rightarrow {P_2} = {3 \over 2}{P_1}$
Putting this value in eqn. (ii) we get
${3 \over 2}{P_1} = {P_1} + {{kx} \over A} \Rightarrow kx = {{{P_1}A} \over 2}$
$x = {{{V_2} - {V_1}} \over A} = {{2{V_1} - {V_1}} \over A} = {{{V_1}} \over A}$
Energy stored in the spring
$ = {1 \over 2}k{x^2} = {1 \over 2}(kx)(x) = {{{P_1}{V_1}} \over 4}$
So, option (a) is correct.
Change in internal energy,
$\Delta U = {f \over 2}({P_2}{V_2} - {P_1}{V_1}) = {3 \over 2}\left( {{3 \over 2}{P_1} 	imes 2{V_1} - {P_1}{V_1}} ight) = 3{P_1}{V_1}$
So, option (b) is correct.
Case II : When ${V_2} = 3{V_1}$ and ${T_2} = 4{T_1}$
From Eqn. (i),
${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}(3{V_1})} \over {4{T_1}}} \Rightarrow {P_2} = {4 \over 3}{P_1}$
$x = {{{V_2} - {V_1}} \over A} = {{2{V_1}} \over A}$
From eqn. (ii),
${4 \over 3}{P_1} = {P_1} + {{kx} \over A} \Rightarrow kx = {{{P_1}A} \over 3}$
Gas is heated very slowly so pressure on the other compartment remains same.
Work done by gas = Work done by gas on atmosphere + Energy stored in spring.
${W_g} = {P_1}Ax + {1 \over 2}k{x^2} = {P_1}(2{V_1}) + {1 \over 2}\left( {{{{P_1}A} \over 3}} ight)\left( {{{2{V_1}} \over A}} ight)$
$ = 2{P_1}{V_1} + {1 \over 3}{P_1}{V_1} = {7 \over 3}{P_1}{V_1}$
So, option (c) is correct.
Heat supplied to the gas,
$\Delta Q = {W_g} + \Delta U$
$ = {7 \over 3}{P_1}{V_1} + {3 \over 2}({P_2}{V_2} - {P_1}{V_1})$
$ = {7 \over 3}{P_1}{V_1} + {3 \over 2}\left( {{4 \over 3}{P_1} 	imes 3{V_1} - {P_1}{V_1}} ight)$
$ = {7 \over 3}{P_1}{V_1} + {9 \over 2}{P_1}{V_1} = {{41} \over 6}{P_1}{V_1}$
So, option (d) is incorrect.
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l0ym34ez/0d47c8c1-8477-411d-b0b9-e45bd44a5661/a3b7ceb0-a7ee-11ec-82ec-9768dcc201cf/file-1l0ym34f0.png?format=png)

Answer

If V
2
= 2V
1
and T
2
= 3T
l
, then the energy stored in the spring is ${1 \over 4}{P_1}{V_1}$

Answer

If V
2
= 2V
1
and T
2
= 3T
1
, then the change in internal energy is $3{P_1}{V_1}$

Answer

If V
2
= 3V
1
and T
2
= 4T
1
, then the work done by the gas is ${7 \over 3}{P_1}{V_1}$

Answer

If V
2
= 3V
1
and T
2
= 4T
1
, then the heat supplied to the gas is ${17 \over 6}{P_1}{V_1}$

Thermodynamics & KTG

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