The work done by 6 moles of helium gas when its temperature increases by 20^{} {C} at constant pressure is (Universal gas constant = 8.31 {J} {mol}^{-1} {K}^{-1} ) [AP EAPCET 23 May 2025 S2]

Correct Answer is : 997.2J997.2 {J}997.2J

Correct Answer is : 997.2J997.2 {J}997.2J

Test Index

Heat and Thermodynamics

Section: Physics

Question : 8 of 78

Marks: +1, -0

20^{\circ} \mathrm{C}

= 8.31 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}

Solution:

Step 1: Use the ideal gas law

We know that

p V = n R T

describes how pressure, volume, and temperature are related for an ideal gas.

Step 2: Set up the formula for work at constant pressure

When pressure is constant, the work (

W

) done by the gas is:

W = p \Delta V

From the ideal gas law, we can find that:

p \Delta V = n R \Delta T

This means the work done is:

W = n R \Delta T

Step 3: Insert the given values

Number of moles,

n = 6

Change in temperature,

\Delta T = 20^{\circ} \mathrm{C}

. (This is the same as 20 K for a temperature difference.)

Gas constant,

R = 8.31 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}

Step 4: Calculate the work

W = 6 \times 8.31 \times 20

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