Thermodynamics

A heat engine is involved with exchange of heat of 1915 J, –40 J, +125 J and –QJ, during one cycle achieving and efficiency of 50.0%. The value of Q is:

A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Assuming the gases to be ideal and the oxygen bond to be rigid, the total internal energy (in units of RT) of the mixture is:
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An engine takes in 5 moles of air at 20°C and 1 atm, and compresses it adiabatically to 1/10th of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be X kJ. The value of X to the nearest integer is:

Initially a gas of diatomic molecules is contained in a cylinder of volume V₁ at a pressure P₁ and temperature 250 K. Assuming that 25% of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature 2000 K, when contained in a volume 2V₁, is given by P₂. The ratio P₂/P₁ is

Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom. The gas is maintained at a temperature of T. The total internal energy, U of a mole of this gas, and the value of γ (=C?/C?) are given, respectively, by:
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For a dimerization reaction, 2A(g) → A₂(g), at 298 K, ∆Uᶱ = -20 kJmol⁻¹, ∆Sᶱ = -30 JK⁻¹mol⁻¹, then the ∆Gᶱ will be J.

In adiabatic process, the density of a diatomic gas becomes 32 times its initial value. The final pressure of the gas is found to be $n$ $$ times the initial pressure. The value of $n$ $$ is:

A litre of dry air at STP expands adiabatically to a volume of 3 litres. If $\gamma =1.40$ $$ , the work done by air is: $\left(3^{1.4}=4.6555\right)$ $\left({}^{}\right)$ [Take air to be an ideal gas]