Question

A sample of oxygen at 40.°C occupies 820. mL. If this sample later occupies 1250 mL at 60.°C
and 1.40 atm, what was its original pressure?

Answer 1

$\tt \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

T₁=40+273=313

T₂=60+273=333

P₁.820/313 = 1.4.1250/333

P₁=2.006 atm