JEE PYQ: Thermodynamics Question 13
Question 13 - 2020 (02 Sep Shift 1)
An engine takes in 5 mole of air at 20$°$C and 1 atm, and compresses it adiabatically to $1/10^{\text{th}}$ 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 $X$ kJ. The value of $X$ to the nearest integer is __________.
Show Answer
Answer: 46
Solution
For adiabatic process, $TV^{\gamma-1} =$ constant
$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$
$T_1 = 20°C + 273 = 293$ K, $V_2 = \frac{V_1}{10}$ and $\gamma = \frac{7}{5}$
$T_1(V_1)^{\gamma-1} = T_2\left(\frac{V_1}{10}\right)^{\gamma-1}$
$293 = T_2\left(\frac{1}{10}\right)^{2/5} \Rightarrow T_2 = 293(10)^{2/5} = 736$ K
$\Delta T = 736 - 293 = 443$ K
$\Delta U = NC_v\Delta T = 5 \times \frac{5}{2} \times 8.3 \times 443 = 46 \times 10^3$ J $= X$ kJ
$\therefore X = 46$
BETA
