Thermodynamics

JEE Main level, numerical answer, medium

51 of 190

Question 51 of 190

An engine takes in 55 moles of air at 20∘C20^{\circ}C and 1atm1atm, and compresses it adiabatically to 1/10th1/{10}^{th} of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be XkJXkJ. The value of XX to the nearest integer is:

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Answer: 46

T1V1γ−1=T2V2γ−1{T}_{1}{V}_{1}^{\gamma -1}={T}_{2}{V}_{2}^{\gamma -1}
T2=T1(V1V2)γ−1{T}_{2}={T}_{1}{\left(\dfrac{{V}_{1}}{{V}_{2}}\right)}^{\gamma -1}
=T1(10)75−1={T}_{1}(10{)}^{\dfrac{7}{5}-1}
T2=T1(10)2/5{T}_{2}={T}_{1}(10{)}^{2/5}
ΔV=52nR;52×5×3[102/5−1](293)\Delta V=\dfrac{5}{2}nR;\dfrac{5}{2}\times 5\times 3\left[{10}^{2/5}-1\right](293)
6256×1.5×293=461440\dfrac{625}{6}\times 1.5\times 293=461440
≈46ks\approx 46ks

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