Question

3. At 10.0°C and 2.00 atm, the volume of the gas is 75.0 mL. If the temperature of the gas increases to

Answer 1

Explanation:

c.

$from \\ \frac{P _{1} V _{1} }{T _{1}} = \frac{P _{2} V _{2}}{T _{2}} \\ \frac{2 \times 75}{(10+273)} = \frac{V _{2} \times 1}{(35+273)} \\V _{2} = \frac{2 \times 75 \times (35+273)}{(10+273) \times 1} \\ V _{2} = 163 \: ml$