Question

A gas has a volume of 800.0 ml at minus 23.00 degrees c and 300.0 torr. what would be the volume of thegas be at 227.0 dgrees c and 600.0 torr of pressure?

Answer 1

Solution:

Initial Pressure, P₁ = 300.0 torr

Initial Volume, V₁ = 800.0 ml

Initial Temperature, T₁ = -23.00°C = -23+273 = 250K

If the conditions are changed,

Changed Pressure, P₂ = 600.0 torr

Changed Temperature, T₂ = 227.0°C = 227+273 = 500K

Changed Volume, V₂ = [To find]

We will apply Combined Gas Law here,

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

$\frac{300.0 * 800.0}{250 } = \frac{600.0 * V_{2 }}{500 }$

$V_{2 } = \frac{300.0 * 800.0 *500}{250 *600.0 }$

$V_{2 } = \frac{300.0 * 800.0 * 2}{600.0 }$

$V_{2 } = \frac{600.0 * 800.0 }{600.0 }$

$V_{2 } = 800.0 \ ml$

The volume of the gas at 227.0°C and 600.0 torr is 800.0 ml