A flexible balloon contains 0.320 mol of hydrogen sulfide gas ({\ m H}_{2}{\ m S

Question

A flexible balloon contains 0.320 mol of hydrogen sulfide gas ({ m H}_{2}{ m S}). Initially the balloon of { m H}_{2}{ m S} has a volume of 6600 cm^3 and a temperature of 21.0^circ C. The { m H}_{2}{ m S} first expands isobarically until the volume doubles. Then it expands adiabatically until the temperature returns to its initial value. Assume that the { m H}_{2}{ m S} may be treated as an ideal gas with C_p = 34.60;{ m J/mol cdot K} and gamma = 4/3.

a) What is the total heat Q supplied to the { m H}_{2} { m S} gas in the process?

b) What is the total change in the internal energy DeltaU of the { m H}_{2}{ m S} gas?

c) What is the total work W done by the { m H}_{2} { m S} gas?

d) What is the final volume V?

Can someone help please?

Explanation / Answer

a)The number of moles of hydrogen sulfide gas is n = 0.320 mol
The initial volume of the balloon is V1 = 6600 cm^3 = 6600 * (10^-2)^3 m^3 = 6600 * 10^-6 m^3
The initial temperature of the balloon is T1 = 21.0 oC
The balloon expands isobarically until the volume doubles,that is,V2 = 2V1
The total heat Q supplied to the gas in the process is
Q = nRT1
where R = 8.314 J/mol/K
b)The total change in the internal energy of the gas is
?U = Q + W + W'
where W is the mechanical work done on a system (measured in joules in SI)
W' is energy added by all other processes
Here,W = P * ?V
where P = 1.013 * 10^5 Pa,?V = V2 - V1 = 2V1 - V1 = V1 and W' = 0 c)The total work done W by the gas is W = Q - W d)The final volume of the gas is equal to its initial volume.

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