Question #4db2c

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Answer 1 · 2017-12-15T03:53:44

(b) $40.5@C$

The initial volume of argon ,V1=1 L,

The initial pressure,P1=720 mmHg

The initial temperature,T1= $20@C=(273+20)K=293K$

The decreased pressure,P2=360 mmHg

The increased volume,V2=2.4 L

The final temperature,T2=?

We know,

$(P1.V1)/(T1)=(P2V2)/(T2)$

=)T2= $(P2V2T1)/(P1V1)=313.5K=40,5@C$