You transfer a sample of gas at 17°C from a volume of 4.81 L and 1.10 atm to a container at 37°C that has a pressure of 1.10 atm. What is the new volume of the gas?

An important thing to notice here is that, unless the question is mistyped, the pressure of the gas is being kept constant.

If you take into account the fact that the number of moles of gas is constant as well, you can say that the change in volume will only depend on the change in temperature.

As you know, when number of moles of gas and pressure are kept constant, volume and temperature have a direct relationship - this is known as Charles' Law.

So, when temperature Increases, the volume increases as well. Likewise, when temperature decreases, the volume decreases as well.

In your case, the temperature increased from #17^@"C"# to #37^@"C"#, which means that you can expect the volume of the gas to increase.

Mathematically, Charles' Law is expressed as

#color(blue)(V_1/T_1 = V_2/T_2)" "#, where

#V_1#, #T_1# - the volume and temperature of the gas at an initial state
#V_2#, #T_2# - the volume and temperature of the gas at a final state

Rearrange to solve for #V_2# - no not forget that the temperature of the gas must be expressed in Kelvin!

#V_2 = T_2/T_1 * V_1#

In your case, this will be equal to

#V_2 = ( (273.15 + 17) color(red)(cancel(color(black)("K"))))/( (273.15 + 37) color(red)(cancel(color(black)("K")))) * "4.81 L"#

#V_2 = "6.641 L"#

Rounded to two sig figs, the number of sig figs you have for the temperatures of the gas, the answer will be

#V_2 = color(green)("6.6 L")#

SIDE NOTE If the pressure of the gas is not constant, you can solve the problem by using the combined gas law equation

#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "#, where

#P_1#, #V_1#, #T_1# - the pressure, volume, and temperature of the gas at an initial state
#P_2#, #V_2#, #T_2# - the pressure, volume, and temperature of the gas at a final state