**Problem:**

A 20 L cylinder containing 6 atm of gas at 27 °C. What would the pressure of the gas be if the gas was heated to 77 °C?

**Solution**

The cylinder's volume remains unchanged while the gas is heated so Gay-Lussac's gas law applies. Gay-Lussac's gas law can be expressed as

P

_{i}/T

_{i}= P

_{f}/T

_{f}

where

P

_{i}and T

_{i}are the initial pressure and absolute temperatures

P

_{f}and T

_{f}are the final pressure and absolute temperature

First, convert the temperatures to absolute temperatures.

T

_{i}= 27 °C = 27 + 273 K = 300 K

T

_{f}= 77 °C = 77 + 273 K = 350 K

Use these values in Gay-Lussac's equation and solve for P

_{f}.

P

_{f}= P

_{i}T

_{f}/T

_{i}

P

_{f}= (6 atm)x(350K)/(300 K)

P

_{f}= 7 atm

**Answer:**

The pressure will increase to 7 atm after heating the gas from 27 °C to 77 °C.