One mole of diatomic gas at 300 K is compressed adiabatically to one-fourth of its volume- If Y -1-5- the work done on gas will beA- 1280 JB- 1610 JC- 2044 JD- 4986 J

The correct option is D 4986 JGiven, γ = 1.5, V_1 = V; V_2 = V4; T_1 = 300 K As, T_1 V^γ 1_1 = T_2 V^γ 1_2 We get,T_2 = T_1 (V_1V_2 )^γ 1 = 300 (4)^0.5 = ...

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Standard XII

Physics

Work Done

One mole of d...

Question

One $mole$ of diatomic gas at $300 K$ is compressed adiabatically to one-fourth of its volume. If $γ = 1.5$ , the work done on gas will be

1280 J

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1610 J

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2044 J

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4986 J

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Solution

The correct option is D $4986 J$
Given,
$γ = 1.5, V_{1} = V; V_{2} = \frac{V}{4}; T_{1} = 300 K$
As, $T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$
We get, $T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{γ - 1} = 300 (4)^{0.5} = 600 K$
for one mole of gas, the work done
$W = ∣ ∣ ∣ \frac{R (T_{1} - T_{2})}{γ - 1} ∣ ∣ ∣ = ∣ ∣ ∣ \frac{8.31 \times (300 - 600)}{0.5} ∣ ∣ ∣$
$= 4986 J$

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One mole of diatomic gas at 300 K is compressed adiabatically to one-fourth of its volume. If γ=1.5, the work done on gas will be

The correct option is D 4986 JGiven, γ=1.5, V1=V; V2=V4; T1=300 K As, T1Vγ−11=T2Vγ−12 We get,T2=T1(V1V2)γ−1=300(4)0.5=600 K for one mole of gas, the work done W=∣∣∣R(T1−T2)γ−1∣∣∣=∣∣∣8.31×(300−600)0.5∣∣∣ =4986 J

One $mole$ of diatomic gas at $300 K$ is compressed adiabatically to one-fourth of its volume. If $γ = 1.5$ , the work done on gas will be