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An ideal gas absorbs `600cal` of heat during expansion from `10L` to `20L` against the constant pressure of `2atm`. Calculate the change in internal enegry. |
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Answer» `w = P(V_(2)-V_(1))` `=2(20-10) L atm = 2xx 10 xx 24.2cal = 484 cal` `:. DeltaU = q - w = 600 - 484 = 116 cal` |
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