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Determine the change in volume if 60ml of gas at 33°C is cooled 15°C


Sagot :

Answer:

V_{2} = 56.47mLV

2

=56.47mL

Explanation:

Use Charles's Law:

when Pressure is constant.

\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} }

T

1

V

1

=

T

2

V

2

arrange:

V_{2} = \frac{V_{1}T_{2}}{T_{1}}V

2

=

T

1

V

1

T

2

GIVEN:

\begin{gathered}V_1 = 60mL\\T_1 = 33+273 = 306K\\T_2 = 15+273 = 288K\end{gathered}

V

1

=60mL

T

1

=33+273=306K

T

2

=15+273=288K

SOLUTION:

V_2=\frac{(60)(288)}{306}V

2

=

306

(60)(288)

V_2 = 56.47 mLV

2

=56.47ml