Question

A sample of argon has a volume of 1.20 L at STP. If the temperature is increased to 26.0°C and the pressure is lowered to 0.800 atm, what will the new volume be, in L?

Answer 1

Key words:-

• Pressure=P
• Volume=V
• Temperature=T

Now

• P1=1atm
• P_2=0.800atm
• V1=1.2L
• V2=L
• T1=273K
• T2=26°C=299K

Apply combined gas law

$\\ \tt\hookrightarrow \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

$\\ \tt\hookrightarrow \dfrac{1(1.2)}{273}=\dfrac{(0.800)V_2}{299}$

$\\ \tt\hookrightarrow 0.004=0.002V_2$

$\\ \tt\hookrightarrow V_2=2L$(Approx)