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Worked Examples · Example 5.3

Q.At 25°C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at a height where temperature is 10°C and volume of the gas is 640 mL.

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Step 1 – Identify the process

All three variables (pp, VV, TT) change and the amount of gas is fixed, so use the combined gas law:

p1V1T1=p2V2T2\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}

Step 2 – Convert temperatures to kelvin

T1=25+273=298 K,T2=10+273=283 KT_1 = 25 + 273 = 298\ \text{K}, \qquad T_2 = 10 + 273 = 283\ \text{K}

Step 3 – List the data

p1=760 mm Hg,V1=600 mL,V2=640 mLp_1 = 760\ \text{mm Hg}, \quad V_1 = 600\ \text{mL}, \quad V_2 = 640\ \text{mL}

Step 4 – Solve for p2p_2

p2=p1×V1V2×T2T1p_2 = p_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} …

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