One gram mol of a diatomic gas Y -1-4 is compressed adiabatically so that its temperature rises from 27- C to 127- C- The work done will beA- 207-85 joulesB- 2078-5 joulesC- 207-85 ergsD- None of the above

The correct option is B 2078.5 joules nbsp;We know, from first law, Δ U = q + w. For adibatic process, q =0. Also Δ U = nC_vdT nbsp; and w = Δ U = nR(γ 1 ...

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Standard XII

Chemistry

Work Done in Reversible Adiabatic Process

One gram mol ...

Question

One gram mol of a diatomic gas $(γ = 1.4)$ is compressed adiabatically so that its temperature rises from $27^{o} C$ to $127^{o} C .$ The work done will be

2078.5 joules

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207.85 joules

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207.85 ergs

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None of the above

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Solution

The correct option is A 2078.5 joules
We know, from first law, $Δ U = q + w$ .
For adibatic process, $q = 0$ .
Also $Δ U$ = $n C_{v} d T$ and w = $Δ U$ = $\frac{n R}{(γ - 1)} (T_{2} - T_{1})$
= $\frac{1 \times R}{(1.4 - 1)} (400 - 300) = 2078.5 J$

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Work Done in Reversible Adiabatic Process

Standard XII Chemistry

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One gram mol of a diatomic gas (γ=1.4) is compressed adiabatically so that its temperature rises from 27oC to 127oC. The work done will be

The correct option is A 2078.5 joules We know, from first law, ΔU=q+w. For adibatic process, q=0. Also ΔU = nCvdT and w = ΔU = nR(γ−1)(T2−T1) = 1×R(1.4−1)(400−300)=2078.5 J

One gram mol of a diatomic gas $(γ = 1.4)$ is compressed adiabatically so that its temperature rises from $27^{o} C$ to $127^{o} C .$ The work done will be

The correct option is A 2078.5 joules
We know, from first law, $Δ U = q + w$ .
For adibatic process, $q = 0$ .
Also $Δ U$ = $n C_{v} d T$ and w = $Δ U$ = $\frac{n R}{(γ - 1)} (T_{2} - T_{1})$
= $\frac{1 \times R}{(1.4 - 1)} (400 - 300) = 2078.5 J$