If quantity of heat 1163.4 J supplied to one mole of nitrogen gas, at room temperature at constant pressure, then the rise in temperature is

Question

If quantity of heat 1163.4 J supplied to one mole of nitrogen gas, at room temperature at constant pressure, then the rise in temperature is (A) 54 K (B) 28 K (C) 65 K (D) 8 K

Tardigrade Admin · Accepted Answer

Heat given to the gas at room temperature and at constant temperature.
Q =

n C_{p} △ T

∴ 1163.4 = 1 \times \frac{7}{2} R \times △ T

(∵ C_{p} = \frac{7}{2}

R for diatomic gas)
or

△ T = \frac{2 \times 1163.4}{7 \times 8.31}

(∵ R = 8.31 J m o l^{- 1} K^{- 1})

or

△ T = 40 K

Q. If quantity of heat 1163.4 J supplied to one mole of nitrogen gas, at room temperature at constant pressure, then the rise in temperature is

54 K

28 K

65 K

8 K

40 K

Heat given to the gas at room temperature and at constant temperature. Q = nCp​△T ∴1163.4=1×27​R×△T (∵Cp​=27​ R for diatomic gas) or △T=7×8.312×1163.4​ (∵R=8.31Jmol−1K−1) or △T=40K

Heat given to the gas at room temperature and at constant temperature.
Q = $n C_{p} △ T$
$∴ 1163.4 = 1 \times \frac{7}{2} R \times △ T$
$(∵ C_{p} = \frac{7}{2}$ R for diatomic gas)
or $△ T = \frac{2 \times 1163.4}{7 \times 8.31}$
$(∵ R = 8.31 J m o l^{- 1} K^{- 1})$
or $△ T = 40 K$