A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas γ = 5 3 and one mole of an ideal diatomic gas γ = 7 5 . Here, γ is the ratio of the specific heats at constant pressure and constant volume of an ideal gas. The gas mixture does a work of 66…

Question

A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas &#947; = 5 3 and one mole of an ideal diatomic gas &#947; = 7 5 . Here, &#947; is the ratio of the specific heats at constant pressure and constant volume of an ideal gas. The gas mixture does a work of 66 Joule when heated at constant pressure. The change in its internal energy is _____ Joule.

Aicharya · Accepted Answer

Change in internal energy for the mixture can be written as, $∆ U = n_{1} {(C_{v})}_{1} ∆ T + n_{2} {(C_{v})}_{2} ∆ T$
$= (n_{1} {(C_{v})}_{1} + n_{2} {(C_{v})}_{2}) ∆ T . . . (i)$
For isobaric process, work done $= P ∆ V$
$= (n_{1} + n_{2}) R ∆ T . . . (ii)$
Divide $(i)$ by $(ii)$ , we get
$\frac{∆ U}{W} = \frac{(n_{1} {(C_{v})}_{1} + n_{2} {(C_{v})}_{2}) ∆ T}{(n_{1} + n_{2}) R ∆ T}$
$\Rightarrow ∆ U = \frac{W}{R} (\frac{(n_{1} {(C_{v})}_{1} + n_{2} {(C_{v})}_{2})}{n_{1} + n_{2}})$
$= \frac{66}{R} \frac{[2 \times \frac{3 R}{2} + 1 \times \frac{5 R}{2}]}{2 + 1}$
$= 121 J$

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A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas γ = 5 3 and…

Solution

A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas &#947; = 5 3 and…

Solution

A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas γ = 5 3 and…