If a quantity of heat 1163.4 joule is supplied to one mole of nitrogen gas, at room temperature at constant pressure, then the rise in temperature is (Given R=8.31 J mole^-1 K^-1 )

Question

Accepted Answer

Correct answer: D. (d) : Heat supplied to the gas at constant pressure is, Q=nC_P T where n is the number of moles, C_p is the molar specific heat at constant pressure, T is the rise in temperature. Since nitrogen is a diatomic gas, so C_P= 72R ∴ Q= 72 nR T or T= 2 Q7 nR= 2 1163.47 1 8.31=40 K

$If a quantity of heat 1163.4 joule is supplied to one mole of nitrogen gas, at room temperature at constant pressure, then the rise in temperature is (Given R = 8.31 J mole^{- 1} K^{- 1})$

Solution

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