First slide

laws

Question

In order to increase the volume of a gas to 3 times at constant pressure at $40^{o} C$ , the final temperature should be

Easy

A

$666^{0} C$
B

$777^{0} C$
C

$555^{0} C$
D

$333^{0} C$

Solution

$\frac{T^{'}}{T} = \frac{3 V}{V} = 3$
$T^{'} = 3 T = 3 [40 + 273] K$
= $3 \times 313 K = 939 K$
T = ${(939 - 273)}^{0} C = 666^{0} C$

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