Question

An ideal gas is confined to a 10.0-L balloon at STP. What is
the new volume of the balloon when it is placed under
800. mmHg at 100°C?

Answer 1

Answer:

$V_2= 13.0L$

Explanation:

Hello.

In this case, we can consider this problem related to ideal gases which are those that are not attracted or repulsed to each other, and can be studied via the ideal gas equation:

$PV=nRT$

Thus, since we have two conditions and the moles of the gas have not changed, we can write:

$\frac{P_1V_1}{T_1}= \frac{P_2V_2}{T_2}$

Because the initial conditions for 10.0 L of the gas are 1 atm and 273.15 K (STP) and the final conditions are 800 mmHg (1.053 atm), so the new volume is:

$V_2=\frac{P_1V_1T_2}{T_1P_2}=\frac{1atm*10.0L*373.15K}{273.15K*1.053atm}\\\\V_2= 13.0L$

Best regards!

Answer 2

When the balloon placed under 800. mmHg, the new volume will be "13.0 L".

Ideal gas equation

According to the question,

Pressure, P₁ = 1 atm

                P₂ = 1.053 atm

Temperature, T₁ = 273.15 K

                       T₂ = 373.15 K

Volume, V₁ = 10.0 L

By using the Ideal gas equation,

→ PV = nRT

or,

→ $\frac{P_1 V_1}{T_2} = \frac{P_2 V_2}{T_2}$

Then,

The new volume be:

→ V₂ = $\frac{P_1 V_1 T_2}{T_1 P_2}$

By substituting the above values,

       = $\frac{1\times 10.0\times 373.15}{273.15\times 1.053}$

       = 13.0 L

Thus the approach above is right.  

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