Question

A gas sample has an original volume of 580 ml when collected at 1.20 atm and 22oC. If a change is made in the gas pressure which causes the volume of the gas sample to become 1.45 liters at 40oC, what is the new pressure?

Answer 1

Answer: The new pressure is 0.509 atm

Explanation:

The combined gas equation is,

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

where,

$P_1$ = initial pressure of gas = 1.20 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 580 ml

$V_2$ = final volume of gas = 1.45 L = 1450 ml     (1L=1000ml)

$T_1$ = initial temperature of gas = $22^0C=(22+273)K=295K$

$T_2$ = final temperature of gas = $40^0C=(40+273)K=313K$

Now put all the given values in the above equation, we get:

$\frac{1.20\times 580}{295}=\frac{P_2\times 1450}{313}$

$P_2=0.509atm$

The new pressure is 0.509 atm