Question

A gas sample is observed to occupy 12.0 L under a pressure of 101.325 kPa (Also 1 ATM) at 27 degree Celsius. what will be the volume of the gas if it is heated to 57 degree Celsius under the same pressure?​

Answer 1

Answer:

V

1

=12.0 L

T_{1} = \text{27°C + 273 = 300 K}T

1

=27°C + 273 = 300 K

T_{2} = \text{57°C + 273 = 330 K}T

2

=57°C + 273 = 330 K

Unknown:

V_{2}V

2

Solution:

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

T

1

V

1

=

T

2

V

2

V_{2} = V_{1} × \frac{T_{2}}{T_{1}}V

2

=V

1

×

T

1

T

2

V_{2} = \text{12.0 L} × \frac{\text{330 K}}{\text{300 K}}V

2

=12.0 L×

300 K

330 K

\boxed{V_{2} = \text{13.2 L}}

V

2

=13.2 L