Question

Find the volume of a gas at STP, if its volume is 80.0 mL at 109 kPa and -12.5°C.​

Answer 1

Answer: The volume when the pressure and temperature has changed is 90.21 mL

Step-by-step explanation:

At STP:

The temperature at this condition is taken as 273 K

The pressure at this condition is taken as 1 atm or 101.3 kPa.

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

where,

`P_1,V_1\text{ and }T_1` are the initial pressure, volume and temperature of the gas

`P_2,V_2\text{ and }T_2` are the final pressure, volume and temperature of the gas

We are given:

`P_1=101.3kPa\\V_1=?mL\\T_1=273K\\P_2=109kPa\\V_2=80.0mL\\T_2=-12.5^oC=[-12.5+273]K=260.5K`

Putting values in above equation, we get:

`(101.3kPa* V_1)/(273K)=(109kPa* 80)/(260.5K)\\\\V_1=(109* 80* 273)/(101.3* 260.5)=90.21mL`

Hence, the volume when the pressure and temperature has changed is 90.21 mL

Answer 2

= 913.84 mL

Step-by-step explanation:

Using the combined gas laws

P1V1/T1 = P2V2/T2

At standard temperature and pressure. the pressure is 10 kPa, while the temperature is 273 K.

V1 = 80.0 mL

P1 = 109 kPa

T1 = -12.5 + 273 = 260.5 K

P2 = 10 kPa

V2 = ?

T2 = 273 K

Therefore;

V2 = P1V1T2/P2T1

= (109 kPa × 80 mL × 273 K)/(10 kPa× 260.5 K)

= 913.84 mL