JEE MAIN - Chemistry (2022 - 29th June Evening Shift - No. 15)
2.2 g of nitrous oxide (N2O) gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causing the compression of the gas from 217.1 mL to 167.75 mL. The change in internal energy of the process, $$\Delta$$U is '$$-$$x' J. The value of 'x' is ________. [nearest integer]
(Given : atomic mass of N = 14 g mol$$-$$1 and of O = 16 g mol$$-$$1. Molar heat capacity of N2O is 100 J K$$-$$1 mol$$-$$1)
Answer
195
Explanation
$\Delta T=-40 \mathrm{~K}$
$$ \begin{aligned} \Delta U &=q+w \\\\ &=\frac{100 \times 2.2}{44}(-40)-(-49.39) \times 10^{-3} \times 101.325 \end{aligned} $$
$$ \begin{aligned} &=-200+5 \\\\ &=-195 \mathrm{~J} \\\\ \mathrm{x}=& 195 \end{aligned} $$
$$ \begin{aligned} \Delta U &=q+w \\\\ &=\frac{100 \times 2.2}{44}(-40)-(-49.39) \times 10^{-3} \times 101.325 \end{aligned} $$
$$ \begin{aligned} &=-200+5 \\\\ &=-195 \mathrm{~J} \\\\ \mathrm{x}=& 195 \end{aligned} $$
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