0.056 kg of Nitrogen is enclosed in a vessel at a temperature of 127$^\circ$C. Th amount of heat required to double the speed of its molecules is ____________ k cal.
Take R = 2 cal mole$-$1 K$-$1)
Answer (integer)
12
Solution
<p>Because the vessel is closed, it will be an isochoric process.</p>
<p>To double the speed, temperature must be 4 times (v $\alpha$$\sqrt{T}$)</p>
<p>So, T<sub>f</sub> = 1600 K, T<sub>i</sub> = 400 K</p>
<p>number of moles are ${{56} \over {28}} = 2$</p>
<p>so Q = nCv $\Delta$T = 2 $\times$ ${5 \over 2}$ $\times$ 2 $\times$ 1200</p>
<p>= 12000 cal = 12 K cal</p>
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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