Question

A 250. mL sample of gas at 1.00 atm and 20.0°C has the temperature increased to 40.0°C and the volume increased to 500. mL. What is the new pressure?
a. 0.374 atm
b. 0.534 atm
c. 2.14 atm
d. 1.87 atm
e. 0.468 atm

Answer 1

Answer : The new pressure is, 0.534 atm

Solution :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 1.00 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 250 ml

$V_2$ = final volume of gas = 500 ml

$T_1$ = initial temperature of gas = $20^oC=273+20=293K$

$T_2$ = final temperature of gas = $40^oC=273+40=313K$

Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{1atm\times 250ml}{293K}=\frac{P_2\times 500ml}{313K}$

$P_2=0.534atm$

Therefore, the new pressure is, 0.534 atm