JEE Main 2020 — Physics Thermodynamics

2020mcqeasy

Two moles of an ideal gas, with $\frac{{C}_{P}}{{C}_{V}}=\frac{5}{3}$ , are mixed with three moles of another ideal gas $\frac{{C}_{P}}{{C}_{V}}=\frac{4}{3}$ . The value of $\frac{{C}_{P}}{{C}_{V}}$ for the mixture is

Official previous-year question

Held on 7 Jan 2020 · Verified 6 Jul 2026.

Options

A
$1.45$
B
$1.50$
C
$1.47$
D
$1.42$

Did you get this right?

Solution

Let $n_1,n_2$ be the number of moles of gases $1$ and $2$ .

We have $\dfrac{C_{p_1}}{C_{v_1}}=\dfrac{5}{3}$ and We know $C_{p_1}=C_{v_1}+R$

$\therefore \dfrac{5C_{v_1}}{3}=C_{v_1}+R,\dfrac{2C_{v_1}}{3}R_1,C_{v_1}=\dfrac{3R}{2}$

$A$ and $C_{p_1}=C_{ v_ 1 }+R=\dfrac{5R}{2}$

Also, $\dfrac{C_{ p_ 2}}{C_{ v_ 2 }}=\dfrac{4}{3}$ and

$C_{p_2}=C_{v_2}+R$

$\therefore \dfrac{4C_{ v_ 2 }}{3}=C_{ v_ 2 }+R, \dfrac{C_{ v_ 2 }}{3}=R,C_{ v_ 2 }=3R$

$C_{ p_ 2}=C_{ v_ 2 }+ R=4R$

Now $C_{ v_ {mix} }=\dfrac{n_1C_{ v_ 1 }+n_2C_{ v_ 2 }}{n_1+n_2}=\dfrac{2\times \dfrac{3R}{2}+3\times 3R}{2+3}$

$=\dfrac{3R+9R}{5}=\dfrac{12R}{5}$

$C_{ p_ {mix} }=\dfrac{n_1C_{ p_ 1 }+n_2C_{ p_ 2 }}{n_1+n_2}=\dfrac{2\times \dfrac{5R}{2}+3\times 4R}{2+3}$

$=\dfrac{17R}{5}$

$l_{mix}=\dfrac{C_{ p_ {mix} }}{C_{ v_ {mix} }}=\dfrac{17}{12}=1.4166=1.42$ .

Your note

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Solution

Let $n_1,n_2$ be the number of moles of gases $1$ and $2$ .

We have $\dfrac{C_{p_1}}{C_{v_1}}=\dfrac{5}{3}$ and We know $C_{p_1}=C_{v_1}+R$

$\therefore \dfrac{5C_{v_1}}{3}=C_{v_1}+R,\dfrac{2C_{v_1}}{3}R_1,C_{v_1}=\dfrac{3R}{2}$

$A$ and $C_{p_1}=C_{ v_ 1 }+R=\dfrac{5R}{2}$

Also, $\dfrac{C_{ p_ 2}}{C_{ v_ 2 }}=\dfrac{4}{3}$ and

$C_{p_2}=C_{v_2}+R$

$\therefore \dfrac{4C_{ v_ 2 }}{3}=C_{ v_ 2 }+R, \dfrac{C_{ v_ 2 }}{3}=R,C_{ v_ 2 }=3R$

$C_{ p_ 2}=C_{ v_ 2 }+ R=4R$

Now $C_{ v_ {mix} }=\dfrac{n_1C_{ v_ 1 }+n_2C_{ v_ 2 }}{n_1+n_2}=\dfrac{2\times \dfrac{3R}{2}+3\times 3R}{2+3}$

$=\dfrac{3R+9R}{5}=\dfrac{12R}{5}$

$C_{ p_ {mix} }=\dfrac{n_1C_{ p_ 1 }+n_2C_{ p_ 2 }}{n_1+n_2}=\dfrac{2\times \dfrac{5R}{2}+3\times 4R}{2+3}$

$=\dfrac{17R}{5}$

$l_{mix}=\dfrac{C_{ p_ {mix} }}{C_{ v_ {mix} }}=\dfrac{17}{12}=1.4166=1.42$ .

JEE Main 2020 — Physics Thermodynamics

Options

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Solution

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JEE Main 2020 — Physics Thermodynamics

Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics