What is the temperature of 350 mL of gas at .85 atm if it has a volume of 920 mL at STP?

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Answer 1 · 2016-05-07T02:47:56

$88.3 K$ $88.3K$

Use the equation:

$\frac{P_{1} \cdot V_{1}}{T_{1}} = \frac{P_{2} \cdot V_{2}}{T_{2}}$ $(P_1 * V_1)/(T_1) = (P_2*V_2)/T_2$

Where $P_{1}$ $P_1$ is the pressure at STP, $T_{1}$ $T_1$ is the temperature at STP and $V_{1}$ $V_1$ is $0.92 L$ $0.92L$ .

Substitute all the values in.

$\frac{(1 a t m) (0.92 L)}{273 K} = \frac{(0.85 a t m) (0.35 L)}{T_{2}}$ $[(1 atm)(0.92L)]/(273K) = [(0.85atm)(0.35L)]/T_2$

$(T_{2}) (0.92 L) = (273 K) (0.85 a t m) (0.35 L)$ $(T_2)(0.92L)= (273K)(0.85atm)(0.35L)$

$(T_{2}) (0.92 L) = 81.22$ $(T_2)(0.92L)= 81.22$

$(T_{2}) = 88.3 K$ $(T_2) = 88.3K$