Question

A 400 ml sample of gas is heated from -20 c to 60

Answer 1

Using p1v1/t1=p2v2/t2
p1=50
p2=225
v1=400ml
v2=?
t1=-20=253k
t2=60=333k
50x400/253=225xv2/333
7.9=0.7xv2
v2=7.9/0.7
v2=11.3ml

Answer 2

Answer : The new volume of gas will be 117 ml.

Explanation:

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 = 50 kPa

$P_2$ = final pressure of gas = 225 kPa

$V_1$ = initial volume of gas = 400 ml

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $-20^0C=(-20+273)K=253K$

$T_2$ = final temperature of gas =  $60^0C=(60+273)K=333K$

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

$\frac{50\times 400}{253K}=\frac{225\times V_2}{333K}$

$V_2=117ml$

Therefore, the new volume of gas will be 117 ml.