Physics problems and solutions, Assignments of Physics

Physics problems and solution about heat and thermodynamics

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2024/2025

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Physics โ€“ Problems and solutions
1. An ideal gas initially at 300 K undergoes an isobaric expansion at 2.5 kPa. If the
volume increases from 1 ๐‘š๐‘š3 to 3 ๐‘š๐‘š3 and 12.5 kJ is transferred to the gas by heat,
what are (a) the change in its internal energy and (b) its final temperature
2. An ideal gas is taken through a quasi-static process described by ๐‘ƒ๐‘ƒ=๐›พ๐›พ๐‘‰๐‘‰2, with
๐›พ๐›พ= 5 ๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘š๐‘š/๐‘š๐‘š6. The gas is expanded to twice its original volume of 1 ๐‘š๐‘š3. How much
work is done on the expanding gas in this process?
3. ๐‘›๐‘› moles of an ideal gas at temperature ๐‘‡๐‘‡1 and volume ๐‘‰๐‘‰1 expand isothermally until
the volume has doubled. In terms of ๐‘›๐‘›, ๐‘‡๐‘‡1 and ๐‘‰๐‘‰1, what are (a) the work done on the
gas, and (b) the heat energy transferred to the gas?
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Physics โ€“ Problems and solutions

1. An ideal gas initially at 300 K undergoes an isobaric expansion at 2.5 kPa. If the volume increases from 1 ๐‘š๐‘š 3 to 3 ๐‘š๐‘š 3 and 12.5 kJ is transferred to the gas by heat, what are (a) the change in its internal energy and (b) its final temperature 2. An ideal gas is taken through a quasi-static process described by ๐‘ƒ๐‘ƒ = ๐›พ๐›พ๐‘‰๐‘‰ 2 , with ๐›พ๐›พ = 5 ๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘š๐‘š/๐‘š๐‘š 6. The gas is expanded to twice its original volume of 1 ๐‘š๐‘š 3. How much work is done on the expanding gas in this process? 3. ๐‘›๐‘› moles of an ideal gas at temperature ๐‘‡๐‘‡ 1 and volume ๐‘‰๐‘‰ 1 expand isothermally until the volume has doubled. In terms of ๐‘›๐‘›, ๐‘‡๐‘‡ 1 and ๐‘‰๐‘‰ 1 , what are (a) the work done on the gas, and (b) the heat energy transferred to the gas?

Solutions:

1.

(a)

๐‘›๐‘› = ๐‘๐‘ ๐‘…๐‘…๐‘‡๐‘‡๐‘‰๐‘‰ = (2.5 ร— 10

(300)(8.31) = 1.003^ ๐‘š๐‘š๐‘š๐‘š๐‘š๐‘š

= (12500)^ โˆ’ (2500)(3 โˆ’ 1)

(b)

For constant ๐‘๐‘,

๐‘‰๐‘‰ 1 ๐‘‡๐‘‡ 1 =^

Work Done

๐‘Š๐‘Š = โˆ’ ๏ฟฝ ๐‘๐‘ ๐‘‘๐‘‘๐‘‰๐‘‰

= โˆ’ ๏ฟฝ ๐‘›๐‘›๐‘…๐‘…๐‘‰๐‘‰๐‘‡๐‘‡ ๐‘‘๐‘‘๐‘‰๐‘‰ ๐‘‰๐‘‰๐‘“๐‘“ ๐‘‰๐‘‰๐‘–๐‘– = โˆ’๐‘›๐‘›๐‘…๐‘…๐‘‡๐‘‡ ๏ฟฝ (^) ๐‘‰๐‘‰^1 ๐‘‘๐‘‘๐‘‰๐‘‰ ๐‘‰๐‘‰๐‘“๐‘“ ๐‘‰๐‘‰๐‘–๐‘– = โˆ’๐‘›๐‘›๐‘…๐‘…๐‘‡๐‘‡ ln ๏ฟฝ๐‘‰๐‘‰๐‘‰๐‘‰๐‘“๐‘“๐‘–๐‘–๏ฟฝ = โˆ’๐‘›๐‘›๐‘…๐‘…๐‘‡๐‘‡ 1 ln(2)

Transferred Heat

Consider ฮ”๐‘ˆ๐‘ˆ = 0

๐‘„๐‘„ = ๐‘›๐‘›๐‘…๐‘…๐‘‡๐‘‡ 1 ln(2)