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Chemistry Question on Thermodynamics

Three moles of an ideal gas are compressed isothermally from 60 L to 20 L using constant pressure of 5 atm. Heat exchange Q for the compression is – ____ Lit. atm.

Answer

As isothermal U=0U = 0 and the process is irreversible:
Q=−W=−[−Pext(V2−V1)]Q = -W = -\left[-P_{\text{ext}}(V_2 - V_1)\right]
Q=5×(20−60)=−200 atm-LQ = 5 \times (20 - 60) = -200 \, \text{atm-L}
Given:
Pext=5 atm,V1=60 L,V2=20 LP_{\text{ext}} = 5 \, \text{atm}, \quad V_1 = 60 \, \text{L}, \quad V_2 = 20 \, \text{L}
Substituting the values:
Q=5×(20−60)=−200 atm-LQ = 5 \times (20 - 60) = -200 \, \text{atm-L}
Thus, the heat exchange for the compression is 200 Lit. atm200 \, \text{Lit. atm}.