JEE MAIN - Chemistry (2019 - 8th April Morning Slot - No. 22)
For silver, Cp(J K–1 mol–1) = 23 +0.01 T. If the temperature (T) of 3 moles of silver is raised from 300
K to 1000 K at 1 atm pressure, the value of $$\Delta H$$ will be close to :
62 KJ
16 KJ
13 KJ
21 KJ
Explanation
Give that,
n = 3
T1 = 300
T2 = 1000
Cp = 23 + 0.01T
We know,
$$\Delta $$H = $$\int\limits_{{T_1}}^{{T_2}} {n{C_p}dT} $$
= $$\int\limits_{300}^{1000} {3\left( {23 + {T \over {100}}} \right)dT} $$
= $$3\left[ {23T + {{{T^2}} \over {200}}} \right]_{300}^{1000}$$
= 3[ 23(1000 - 300 + $${{1 \over {200}}}$$((1000)2 - (300)2)]
= 3[ 23 $$ \times $$ 700 + $${{700 \times 1300} \over {200}}$$ ]
= 61950 J
= 61.95 kJ
$$ \simeq $$ 62 kJ
n = 3
T1 = 300
T2 = 1000
Cp = 23 + 0.01T
We know,
$$\Delta $$H = $$\int\limits_{{T_1}}^{{T_2}} {n{C_p}dT} $$
= $$\int\limits_{300}^{1000} {3\left( {23 + {T \over {100}}} \right)dT} $$
= $$3\left[ {23T + {{{T^2}} \over {200}}} \right]_{300}^{1000}$$
= 3[ 23(1000 - 300 + $${{1 \over {200}}}$$((1000)2 - (300)2)]
= 3[ 23 $$ \times $$ 700 + $${{700 \times 1300} \over {200}}$$ ]
= 61950 J
= 61.95 kJ
$$ \simeq $$ 62 kJ
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