2.2 g of nitrous oxide $\left(\mathrm{N}_{2} \mathrm{O}\right)$ 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, $\Delta \mathrm{U}$ is ' -x ' J . The value of ' x ' is \_\_\_\_ . [nearest integer] (Given: atomic mass of $\mathrm{N}=14 \mathrm{~g} \mathrm{~mol}^{-1}$ and of O $=16 \mathrm{~g} \mathrm{~mol}^{-1}$. Molar heat capacity of $\mathrm{N}_{2} \mathrm{O}$ is $100 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$)

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Chemical Thermodynamics

2.2 g of nitrous oxide

(N_{2} O)

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,

Δ 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

N_{2} O

100 JK^{- 1} mol^{- 1}

)

Answer: 195

Solution

For a constant pressure process:

Step 1: Calculate moles of

N_{2} O

Molar mass = $2 (14) + 16 = 44$ g/mol
$n = \frac{2.2}{44} = 0.05$ mol

Step 2: Calculate

Δ H

at constant pressure:

$Δ T = 270 - 310 = - 40$ K
$Δ H = n C_{p} Δ T = 0.05 \times 100 \times (- 40) = - 200$ J

Step 3: Calculate work done:

$w = - P_{e x t} Δ V = - 1 \times \frac{( 167.75 - 217.1 )}{1000} \times 101.3$
$w = - 1 \times (- 0.04935) \times 101.3 \approx + 5$ J

Step 4: Apply first law:

$Δ U = q_{p} + w = Δ H + w = - 200 + 5 = - 195$ J

Therefore,

x = 195

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Integer answer

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JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Chemical Thermodynamics

Solution

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