One mole of an ideal gas expands isothermally and reversibly from 100 atm to 10 atm at 300K. The largest mass that can be lifted through a height of 1 meter in this expansion (in $k g$ ) is (Take R = 25/3 J/molK, 1/12 latm/molK, g = $10 m / s^{2}$ )

250.00

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250.0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

250

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Solution

Work done by the system $= - n R T l o g \frac{P_{1}}{P_{2}}$
$= - 25 / 3 \times 300 J$
$2500 = M \times g \times Δ h$
$M = 250$ kg

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10 mole of ideal gas expands isothermally and reversibly from a pressure of 10 atm to 1 atm at 300 K. What is the heavisest mass which can lifted through a height of 100 meter by this gas?

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?

10

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

10 a t m

1 a t m

300 K

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

100

meter?

10 mole of ideal gas expands isothermally and reversibly from a pressure of 10 atm to 1 atm at 300 K. What is the heavisest mass which can lifted through a height of 100 meter by this gas?

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One mole of an ideal gas expands isothermally and reversibly from 100 atm to 10 atm at 300K. The largest mass that can be lifted through a height of 1 meter in this expansion (in kg) is (Take R = 25/3 J/molK, 1/12 latm/molK, g = 10 m/s2)

Work done by the system=−nRTlogP1P2 =−25/3×300 J 2500=M×g×Δh M=250 kg

One mole of an ideal gas expands isothermally and reversibly from 100 atm to 10 atm at 300K. The largest mass that can be lifted through a height of 1 meter in this expansion (in $k g$ ) is (Take R = 25/3 J/molK, 1/12 latm/molK, g = $10 m / s^{2}$ )

Work done by the system $= - n R T l o g \frac{P_{1}}{P_{2}}$
$= - 25 / 3 \times 300 J$
$2500 = M \times g \times Δ h$
$M = 250$ kg