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  • An ideal gas undergoes a reversible isobaric expansion at a constant pressure of . T

An ideal gas undergoes a reversible isobaric expansion at a constant pressure of 2.6 \mathrm{~atm}. The gas absorbs 1200 \mathrm{~J} of heat and its temperature increases by 50^{\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 \mathrm{28 \mathrm{~J} / \mathrm{K}.}
 

Option: 1

1400 \mathrm{~J}
 


Option: 2

1500 \mathrm{~J}
 


Option: 3

1590 \mathrm{~J}

 


Option: 4

2003 \mathrm{~J}


Answers (1)

best_answer

Given data:

Heat absorbed \mathrm{(Q)=1200 \mathrm{~J}}

Change in temperature \mathrm{(\Delta T)=50^{\circ} \mathrm{C}=50 \mathrm{~K}}

Heat capacity at constant pressure \mathrm{\left(C_p\right)=28 \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=(28 \mathrm{~J} / \mathrm{K}) \cdot(50 \mathrm{~K})=1400 \mathrm{~J} }

Therefore, the change in heat enthalpy of the gas is \mathrm{ 1400 \mathrm{~J} }.

So, the correct option is 1.

Posted by

Rakesh

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