In an isochoric process, if $ T_{1}=27{ }^{\circ} \mathrm{C} $ and $ T_{2}=127{ }^{\circ} \mathrm{C} $, then $ \frac{p_{1}}{p_{2}} $ will be equal to

Question

In an isochoric process, if $ T_{1}=27{ }^{\circ} \mathrm{C} $ and $ T_{2}=127{ }^{\circ} \mathrm{C} $, then $ \frac{p_{1}}{p_{2}} $ will be equal to, $ \frac{9}{59} $, $ \frac{2}{3} $, $ \frac{3}{4} $, $ \frac{1}{3} $

MARKS App · Accepted Answer

In an isochoric process, volume of the gas remains constant. As per ideal gas equation, p V = n R T . Thus, at constant volume, p ∝ T . ⇒ p 1 p 2 = T 1 T 2 ⇒ p 1 p 2 = 27 + 273 127 + 273 ⇒ p 1 p 2 = 300 400 = 3 4 .

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