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First Law of Thermodynamics

1 mole of an ...

Question

$1 m o l e$ of an ideal gas $A (C_{v, m} = 3 R)$ and $2 m o l e$ of an ideal gas $B$ are $(C_{V, m} = \frac{3}{2} R)$ taken in a container and expanded reversible and adiabatically from $1 l i t r e t o 4 l i t r e$ starting from initial temperature of $320 K$ . $Δ E$ or $Δ U$ for the process is:

- 240 R

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240 R

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480 R

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- 960 R

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Solution

The correct option is D $- 960 R$
$A v e r a g e C_{v, m} = \frac{n_{1} C_{v, m_{1}} + n_{2} C_{v, m_{2}}}{n_{1} + n_{2}} = 2 R$
For adiabatic process $d U = d W$
$\frac{d T}{T} = - \frac{R}{C_{v, m}} (\frac{d V}{V}) n_{1} C_{V, m_{1}} d T + n_{2} C_{v, m_{2}} d T = - (n_{1} R T + n_{2} R T) \times \frac{d V}{V} I n \frac{T_{2}}{T_{1}} = - \frac{1}{2} I n (\frac{V_{2}}{V_{1}}) \Rightarrow T_{2} = 320 \times {(\frac{1}{4})}^{\frac{1}{2}} = 160 K Δ U = (n_{1} C_{V, m_{1}} + n_{2} C_{V, m_{2}}) Δ T = - 960 R$

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