Question

6dm3 of a dry gas at temperature of 273°C has pressure of 380mm of Hg. Find the volume of gas at
STP
​

Answer 1

Given -

=> At a Temperature of 273 °C and Pressure of 380 mm of Hg a certain mass of dry gas has a volume of 6 dm3 .

Converting temperature to  kelvins -

=> We know that -

K = °C + 273

=> Thus,

K = 273 °C + 273

K = 546 K

=> So, the initial temperature of the gas in kelvins was 546 K .

To find -

=> The volume of the dry gas at STP (Standard Temperature and Pressure)

Some things that you should know -

=> STP conditions are -

• 273 K ( 0 °C) (Temperature)
• 760 mm of Hg (Pressure)

Concepts Used Here -

=> The concept of Gas Laws is used here -

$\dfrac{P1*V1}{T1} = \dfrac{P2*V2}{T2}$

 Where -

P1 - Initial Pressure

P2 - Final Pressure

V1 - Initial Volume

V2 - Final Volume

T1 - Initial Temperature

T2 - Final Temperature

Method (Using gas laws) -

Listing the values accordingly -

P1 = 380 mm of Hg      P2 = 760 mm of Hg

V1 = 6 dm3                   V2 = ?

T1 = 546 K                    T2 = 273 K

Putting the values into the equation-

➼ $\dfrac{380*6}{546} =\dfrac{760*V2}{273}$

➼ $\dfrac{380*6*273}{546*760} = V2$

➼ $\dfrac{1*6*1}{2*2} = V2$

➼ $\dfrac{6}{4} = V2$

➼ V2 = $\dfrac{3}{2}$

➼ V2 = 1.5 dm3

Answer -

=> Thus, the volume of that mass of gas at STP will be 1.5 dm3 or 1.5 litres .

More to know -

=> NTP means Normal Temperature and Pressure, NTP is the same as RTP (Room Temperature and Pressure) -

• 25 °C or 298 K Temperature
• 760 mm of Hg or 1 atm Pressure