A certain gas begins at #"10 atm"#, #"300 K"#, and #"1 L"#. If it expands in a closed container to #"5 atm"# and #"4 L"#, and the heat capacity is #"50 J/"^@ "C"#, what is the change in enthalpy?

Assuming the gas is ideal, you're going from #(P_1 = "10 atm", V_1 = "1 L")# to #(P_2 = "5 atm", V_2 = "4 L")#. Since the pressure halved and the volume quadrupled, i.e.

#P -> 1/2P#
#V -> 4V#,

that means

#[T = (PV)/(nR)] -> [(P/2 cdot 4V)/(nR) = (2PV)/(nR) = 2T]#

So the new temperature is #"600 K"#. WHICH heat capacity was given, I wouldn't know... #C_V# or #C_P#? Well... presumably #C_P#, if you were asked about enthalpy, but not necessarily...

#C_P = ((delH)/(delT))_P#

so

#DeltaH = int_(T_1)^(T_2) C_P dT#

Assuming that #C_P# is constant in the temperature range, #C_P(T) ~~ C_P# and

#color(blue)(DeltaH) ~~ C_P int_(T_1)^(T_2) dT#

#= "50 J/"^@ "C" cdot (T_2 - T_1)#

#= "50 J/"^@ "C" cdot ("600 K" - "300 K")#

#= "50 J/"cancel"K" cdot 300 cancel"K"#

#=# #"15000 J"#

#=# #color(blue)("15 kJ")#

If you were given #C_V#, then we would need to know the mols of gas.

#n = (PV)/(RT) = "constant..."#

#= ("10 atm" cdot "1 L")/("0.082057 L"cdot"atm/mol"cdot"K" cdot "300 K") = "0.4062 mols"#

And with that, if you were given #C_V#, then

#color(blue)(DeltaH) ~~ (C_V + nR) int_(T_1)^(T_2) dT#

#= ("50 J/"^@ "C" + nR) cdot (T_2 - T_1)#

#= ("50 J/"cancel"K" + 0.4062 cancel"mols" cdot "8.314472 J/"cancel"mol"cdotcancel"K") cdot (600 cancel"K" - 300 cancel"K")#

#=# #"16013 J"#

#=# #color(blue)("16 kJ")#