Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A certain amount of an ideal gas occupies a volume of 16 liters at a pressure of 6 atmospheres (atm) and a temperature of $300 \mathrm{~K}$ . The gas is allowed to expand adiabatically to a final volume of 64 liters. Calculate: Change in enthalpy

Option: 1
$\mathrm{278 \mathrm{~L}-atm}$

Option: 2
$\mathrm{288 L-atm}$

Option: 3
$\mathrm{300\: L-atm}$

Option: 4
$\mathrm{430\: L-atm}$

Answers (1)

best_answer

Step 1: Use the ideal gas law to relate the initial and final conditions.

$\mathrm{ P V=\eta R T }$
Step 2: Calculate the initial and final pressures using the ideal gas law.

$\mathrm{ P_1=\frac{n R}{V_1} T_1 }$

$\mathrm{ P_2=\frac{n R}{V_2} T_2 }$

Step 3: Calculate the change in internal energy using the formula for ideal gases.

$\mathrm{ \Delta U=\frac{5}{2} n R\left(T_2-T_1\right) }$

Step 4: Substitute the given values and calculate the change in internal energy.

$\mathrm{ \Delta U=\frac{5}{2} \times(1 \mathrm{~mol}) \times(8.314 \mathrm{~L}-\mathrm{atm} / \mathrm{mol}-\mathrm{K}) \times(300 \mathrm{~K}-300 \mathrm{~K}) }$

The change in internal energy $\mathrm{ (\Delta U) }$ for an adiabatic process is zero.

Step 1: Use the formula for change in heat enthalpy $\mathrm{ (\Delta H) }$ for an ideal gas at constant pressure.

$\mathrm{ \Delta H=\Delta U+P \Delta V }$

Step 2: Calculate the change in volume $\mathrm{ (\Delta V) }$ using the initial and final volumes.

$\mathrm{ \Delta V=V_2-V_1 }$

Step 3: Calculate the change in heat enthalpy using the calculated values.

$\mathrm{ \Delta H=P_2 \Delta V }$

Step 4: Substitute the given values and calculate the change in heat enthalpy.

$\mathrm{ \Delta H=(6 \mathrm{~atm}) \times(64 \mathrm{~L}-16 \mathrm{~L}) }$

Calculating the numerical value of the change in heat enthalpy:

$\mathrm{ \Delta H=288 \mathrm{~L}-\mathrm{atm} }$

So,Correct option is 2 .

Posted by

rishi.raj

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Trending Articles/News

anam-khan

JEE Main 2026: IIT Delhi’s CSE, EE most opted; last 2 year’s closing ranks for all categories

anam-khan

Must provide CBSE 11th registration numbers to JEE Mains aspirants, board tells schools

anam-khan

JEE Main 2026: Engineering at NIT Surathkal tougher for insiders, show BTech cut-offs

Latest Question