3 moles of a monoatomic gas requires 45 cal heat for 5°C rise of temperature at constant volume, then heat required for 5 moles of same gas under constant pressure for 10°C rise of temperature is (R=2 cal/mole/°k)

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200 cal

400 cal

100 cal

250 cal

answer is D.

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Detailed Solution

n = 3 molesfor monoatiomic gas QV = 45cal ΔT = 500CR = 2cal/mole–kCp = Cv + R = 3 + 2 = 5cal/mole - kNow n = 5 mole, ΔT = 10°CQp

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