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Heat and Thermodynamics question

2005 · Shift 0 · Q149

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JEE Main
Physics
Heat and Thermodynamics
2005 · Shift 0 · Q149

Heat and Thermodynamics question

2005 · Shift 0 · Q149

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1

A gaseous mixture consists of

16g

of helium and

16g

of oxygen. The ratio

{{Cp} \over {{C_v}}}

of the mixture is

A
$1.62$
B
$1.59$
C
$1.54$
D
$1.4$

View written solutionFree

Correct answer: A

Find moles of each gas

For helium: $n_{\text{He}}=\frac{16}{4}=4\ \text{mol}$
For oxygen: $n_{\text{O}_2}=\frac{16}{32}=0.5\ \text{mol}$

Use molar heat capacities at constant volume

Helium is a monoatomic gas: $C_{V,\text{He}}=\frac{3R}{2}, \qquad C_{P,\text{He}}=\frac{5R}{2}$
Oxygen is a diatomic gas (at ordinary temperature): $C_{V,\text{O}_2}=\frac{5R}{2}, \qquad C_{P,\text{O}_2}=\frac{7R}{2}$

Calculate total $C_V$ of the mixture

$C_V=n_{\text{He}}\left(\frac{3R}{2}\right)+n_{\text{O}_2}\left(\frac{5R}{2}\right)$

$C_V=4\cdot \frac{3R}{2}+0.5\cdot \frac{5R}{2}$

$C_V=6R+1.25R=7.25R$

Calculate total $C_P$ of the mixture

$C_P=n_{\text{He}}\left(\frac{5R}{2}\right)+n_{\text{O}_2}\left(\frac{7R}{2}\right)$

$C_P=4\cdot \frac{5R}{2}+0.5\cdot \frac{7R}{2}$

$C_P=10R+1.75R=11.75R$

Find the ratio $\gamma=\dfrac{C_P}{C_V}$

$\gamma=\frac{11.75R}{7.25R}=\frac{11.75}{7.25}$

$\gamma\approx 1.621$

So, $\boxed{\frac{C_P}{C_V}\approx 1.62}$

Option check

A: $1.62$ ✅
B: $1.59$ ❌
C: $1.54$ ❌
D: $1.4$ ❌

Therefore, the correct answer is A.

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