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An ideal gas undergoes a reversible isobaric expansion at a constant pressure of 2.8 \mathrm{~atm}. The gas absorbs 1600 \mathrm{~J} of heat and its temperature increases by 60^{\circ} \mathrm{C}. Calculate the change in heat enthalpy of the gas. Given the heat capacity at constant pressure \mathrm{\left(C_p\right)} for the gas is 26 \mathrm{~J} / \mathrm{K}.
 

Option: 1

1600 \mathrm{~J}


 


Option: 2

1560 \mathrm{~J}
 


Option: 3

1450 \mathrm{~J}
 


Option: 4

440 \mathrm{~J} / \mathrm{K}


Answers (1)

best_answer

Given data:
Heat absorbed \mathrm{(Q)=1600 \mathrm{~J}}

Change in temperature \mathrm{(\Delta T)=60^{\circ} \mathrm{C}=60 \mathrm{~K}}

Heat capacity at constant pressure \mathrm{\left(C_p\right)=26 \mathrm{~J} / \mathrm{K}}

The change in heat enthalpy \mathrm{(\Delta H)} for an isobaric process is given by:

\mathrm{ \Delta H=C_p \cdot \Delta T=(26 \mathrm{~J} / \mathrm{K}) \cdot(60 \mathrm{~K})=1560 \mathrm{~J} }

Therefore, the change in heat enthalpy of the gas is \mathrm{ 1560 \mathrm{~J}. }

So, correct option is 2

Posted by

manish painkra

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