$10$ moles of ideal gas expand isothermally and reversibly from a pressure of $10 a t m$ to $1 a t m$ at $300 K$ . What is the largest mass which can lifted through a height of $100$ meter?

31842 k g

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58.55 k g

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342.58 k g

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None of these

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Solution

The correct option is B $58.55 k g$
Workdone $= - n R T ln \frac{P_{1}}{P_{2}} =$ Change in potential energy of the mass.

$∴ - 10 \times 8.314 \times 300 \times ln (\frac{1}{10}) = m \times 9.81 \times 100$

$∴ m = \frac{57431}{981} = 58.55$ $k g$

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Q. 10 mole of ideal gas expand isothermally and reversibly from a pressure of 10 atm to 1 atm at 300 K. What is the largest mass which can be lifted through a height of 100 meter?

5

moles of an ideal gas expand isothermally and reversibly from a pressure of

10 a t m

2 a t m

300 K

. What is the largest mass which can be lifted through a height of

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10 moles of ideal gas expand isothermally and reversibly from a pressure of 10 atm to 1 atm at 300 K. What is the largest mass which can lifted through a height of 100 meter?

The correct option is B 58.55 kgWorkdone =−nRTlnP1P2= Change in potential energy of the mass. ∴−10×8.314×300×ln(110)=m×9.81×100 ∴m=57431981=58.55 kg

$10$ moles of ideal gas expand isothermally and reversibly from a pressure of $10 a t m$ to $1 a t m$ at $300 K$ . What is the largest mass which can lifted through a height of $100$ meter?

The correct option is B $58.55 k g$
Workdone $= - n R T ln \frac{P_{1}}{P_{2}} =$ Change in potential energy of the mass.

$∴ - 10 \times 8.314 \times 300 \times ln (\frac{1}{10}) = m \times 9.81 \times 100$

$∴ m = \frac{57431}{981} = 58.55$ $k g$