The volume of a gas is reduced adiabatically to 14of its volume at 270C. If the value of γ=1.4 then the new temperature will be

T1T2=V2V1r−1300T2=141.4−1T2=30040.4k

The volume of a gas is reduced adiabatically to 14of its volume at 270C. If the value of γ=1.4 then the new temperature will be

T1T2=V2V1r−1300T2=141.4−1T2=30040.4k

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The volume of a gas is reduced adiabatically to $\frac{1}{4}$ of its volume at $27^{0} C$ . If the value of $γ = 1.4$ then the new temperature will be

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$300 \times 4 K$

$150 \times 4^{0.2} K$

$300 \times 4^{0.4} K$

$350 \times 4^{0.4} K$

answer is C.

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Detailed Solution

$\frac{T_{1}}{T_{2}} = {(\frac{V_{2}}{V_{1}})}^{r - 1}$

$\frac{300}{T_{2}} = {(\frac{1}{4})}^{1.4 - 1}$

$T_{2} = (300) 4^{0.4} k$

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The volume of a gas is reduced adiabatically to 14of its volume at 270C. If the value of γ=1.4 then the new temperature will be

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The volume of a gas is reduced adiabatically to $\frac{1}{4}$ of its volume at $27^{0} C$ . If the value of $γ = 1.4$ then the new temperature will be