Question

A gas occupies a volume of 180 mL at 35 °C and 95.9 kPa. What is the volume of the gas at conditions of STP?

Answer 1

151 mL is the correct answer to the given question .

Step-by-step explanation:

We know that

`PV =n RT`

Where P =pressure ,V=volume and T=Temperature

Given

P=95.9 kPa.

V=
`180 * 10 ^(-3)`

R=25/3

T=273 + 35 =308k

Putting these value into the equation we get

`95.9\ * 180\ *\ 10^(-3) \ =\ n * (25)/(3) * 308`

n=
`6.72 * 10^(-3)`

Now using the equation

`n= \ (V)/(22.4)`

`6.72 * 10^(-3) =(V)/(22.4)\\ V\ =\ 150.6mL`

This can be written as 151mL

Answer 2

the volume of the gas at conditions of STP = 151.04998 ml

Step-by-step explanation:

Data given:

V1 = 180 ml

T1 = 35°C or 273.15 + 35 = 308.15 K

P1 = 95.9 KPa

V2 =?

We know that at STP

P2 = 1 atm or 101.3 KPa

T2 = 273.15 K

At STP the pressure is 1 atm and the temperature is 273.15 K

applying Gas Law:

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

putting the values in the equation of Gas Law:

`V_2=(P_1V_1T_2)/(T_1P_2)`

V_2 =
`(95.9*180*273.15)/(308.15*101.3)`

V2 = 151.04998

therefore, V2 = 151.04998 ml