Question #4db2c

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Answer 1 · 2017-12-15T03:53:44

(b) $40.5 \circ C$ $40.5@C$

The initial volume of argon ,V1=1 L,

The initial pressure,P1=720 mmHg

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

The decreased pressure,P2=360 mmHg

The increased volume,V2=2.4 L

The final temperature,T2=?

We know,

$\frac{P 1 . V 1}{T 1} = \frac{P 2 V 2}{T 2}$ $(P1.V1)/(T1)=(P2V2)/(T2)$

=)T2= $\frac{P 2 V 2 T 1}{P 1 V 1} = 313.5 K = 40, 5 \circ C$ $(P2V2T1)/(P1V1)=313.5K=40,5@C$