A canister containing air has a volume of $85$ $85$ $c m^{3}$ $cm^3$ and a pressure of $1.45$ $1.45$ $a t m$ $atm$ when the temperature is $310$ $310$ $K$ $K$ . What is the pressure when the volume is increased to $180$ $180$ $c m^{3}$ $cm^3$ and the temperature is reduced to $280$ $280$ $K$ $K$ ?

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Answer 1 · 2017-03-02T11:57:03

The pressure will fall to $p_{2} \approx 0.523$ $p_2~~0.523$ $a t m$ $atm$

According to the ideal gas equation we know that:

$\frac{p V}{T} = c o n s t$ $(pV)/T=const$

so if we write that:

$p_{1} = 1 a t m$ $p_1=1atm$

$v_{1} = 85 c m^{3}$ $v_1=85cm^3$

$T_{1} = 310 K$ $T_1=310K$

After the change we have:

$v_{2} = 180 c m^{3}$ $v_2=180cm^3$

$T_{2} = 280 K$ $T_2=280K$

We have to calculate the new pressure $p_{2}$ $p_2$

Using the equation stated at the beginning we have:

$\frac{p_{1} \cdot v_{1}}{T_{1}} = \frac{p_{2} \cdot v_{2}}{T_{2}}$ $(p_1*v_1)/T_1=(p_2*v_2)/T_2$

If we transform the formula we get:

$p_{2} = \frac{p_{1} \cdot v_{1} \cdot T_{2}}{T_{1} \cdot v_{2}}$ $p_2=(p_1*v_1*T_2)/(T_1*v_2)$

If we substitute the given data we get:

$p_{2} = \frac{1 \cdot 85 \cdot 310}{180 \cdot 280} = \frac{2635}{5040} \approx 0.523$ $p_2=(1*85*310)/(180*280)=2635/5040~~0.523$